NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

Q: What are the important topics covered in NCERT Exemplar Class 11 Maths Chapter 15?

NCERT Exemplar Class 11 Maths Chapter 15, "Statistics," focuses on measures of dispersion, including range, mean deviation, variance, and standard deviation, with a focus on both ungrouped and grouped data, and frequency distributions.

Q: Why is Statistics important in Class 11 Mathematics?

Statistics is important in Class 11 Mathematics because it provides tools for analyzing and interpreting data, which is crucial for understanding various real-world phenomena and making informed decisions. It helps in understanding data distribution, measures of central tendency (mean, median, mode), and dispersion (variance, standard deviation).

Q: What is the difference between variance and standard deviation in Statistics?

In statistics, variance measures the spread of data points around the mean, while standard deviation is the square root of the variance and provides a measure of spread in the same units as the original data.

Q: How to prepare for Class 11 Maths Statistics chapter effectively?

To effectively prepare for Class 11 Maths Statistics, focus on understanding the concepts, practice various problems, and revise regularly, using NCERT solutions and other resources to reinforce your knowledge.

NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

Edited By Komal Miglani | Updated on Mar 31, 2025 12:20 AM IST

Statistics class 11 NCERT Exemplar solutions is a lesson that students deal with in their day-to-day life, whether they know or not their relevance to Statistics. When a teacher wants to find the average marks in his subject of class, he will use the concept of mean. If he wants to find the marks most often occurring, he will use the concept of mode, and similarly, to find the middle marks, he will use the concept of median. This is how we can use the concepts of statistics in sports to analyse the performance of players, in forecasting to predict the weather by analysing previous data etc.
The NCERT Exemplar Class 11 Maths chapter 15 solutions will help you to understand these basic concepts with the help of different examples based on different scenarios. Students can download the PDF for Class 11 Maths NCERT Exemplar Solutions Chapter 15, which consists of all the important questions. You can cover all the important topics and subtopics like mean, median, mode, variance, and standard deviation. You can also get your queries solved by referring to the solutions solved step by step in an easily comprehensible format. For other chapter solutions, you can refer to NCERT Class 11 Maths Solutions for Other Chapters.

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Class 11 Maths Chapter 15 exemplar solutions Exercise: 15.1
Page number: 278-283
Total questions: 46

Question:1

Find the mean deviation about the mean of the distribution:
$\begin{array}{llllll} Size & 20 & 21 & 22 & 23 & 24 \\ Frequency & 6 & 4 & 5 & 1 & 4 \end{array}$

Answer:

We have to find the mean deviation from the mean of the distribution in this question.

Let us make a table of the given data and fill up the other columns after the calculations.

$\begin{array}{lll} Size (x_{i}) & Frequency (f_{i}) & f_{i} x_{i} \\ 20 & 6 & 120 \\ 21 & 4 & 84 \\ 22 & 5 & 110 \\ 23 & 1 & 23 \\ 24 & 4 & 96 \\ Total & 20 & 433 \end{array}$
$Here, mean \overset{―}{X} = \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{433}{20} = 21.65$

So the above table can be rewritten as
$\begin{array}{lllll} Size (x_{i}) & Frequency (f_{i}) & f_{i} x_{i} & d_{i} = | x_{i} - \bar{x} | & f_{i} d_{i} \\ 20 & 6 & 120 & 1.65 & 9.90 \\ 21 & 4 & 84 & 0.65 & 2.60 \\ 22 & 5 & 110 & 0.35 & 1.75 \\ 23 & 1 & 23 & 1.35 & 1.35 \\ 24 & 4 & 96 & 2.35 & 9.40 \\ Total & 20 & 433 & 6.35 & 25.00 \end{array}$

Hence, mean deviation becomes= $\frac{Σ f_{i} (x_{i} - \bar{x_{i}})}{Σ f_{i}} = \frac{25}{20} = 1.25$

Question:2

Find the mean deviation about the median of the following distribution:

$\begin{array}{llllll} Marks obtained & 10 & 11 & 12 & 14 & 15 \\ No. of students & 2 & 3 & 8 & 3 & 4 \end{array}$

Answer:

The data distribution is given in the question.

We have to find the mean deviation from the median of the distribution in this question.

Let us make a table of the given data and fill up the other columns after calculations

$\begin{array}{lll} Marks obtained (x_{i}) & Number of students (f_{i}) & Cumulative frequency \\ 10 & 2 & 2 \\ 11 & 3 & 5 \\ 12 & 8 & 13 \\ 14 & 3 & 16 \\ 15 & 4 & 20 \\ Total & 20 \end{array}$

$Now here N=20 which is even .$

Here,median $M = \frac{1}{2} [{(\frac{N}{2})}^{t h} o b s e r v a t i o n + {(\frac{N}{2} + 1)}^{t h} o b s e r v a t i o n]$

$M = \frac{1}{2} [{(\frac{20}{2})}^{t h} o b s e r v a t i o n + {(\frac{20}{2} + 1)}^{t h} o b s e r v a t i o n]$

$M = \frac{1}{2} [10^{t h} o b s e r v a t i o n + 11^{t h} o b s e r v a t i o n]$

$Both these observations lie in cumulative frequency 13, for which the corresponding observation is 12$

$M = \frac{1}{2} [12 + 12] = 12$

$So the above updated table is as shown below$

$\begin{array}{lllll} \begin{matrix} Marks obtained \\ (x_{i}) \end{matrix} & \begin{matrix} Number of \\ students (f_{i}) \end{matrix} & \begin{matrix} Cumulative \\ frequency \end{matrix} & d_{i} = | x_{i} - M | & f_{i} d_{i} \\ 10 & 2 & 2 & 2 & 4 \\ 11 & 3 & 5 & 1 & 3 \\ 12 & 8 & 13 & 0 & 0 \\ 14 & 3 & 16 & 2 & 6 \\ 15 & 4 & 20 & 3 & 12 \\ Total & 20 & 6.35 & 25 \end{array}$

Hence, mean deviation becomes $= \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{25}{20} = 1.25$

Question:3

Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.

Answer:

A set of first n natural numbers when n is an odd number is given

We have to find the mean deviation from the mean

We know that the first natural numbers are 1, 2, 3 ……., n.

It is given that n is an odd number.

So, the mean is

$\overset{―}{x} = \frac{1 + 2 + 3 + \dots . . + n}{n} = \frac{\frac{n (n + 1)}{2}}{n} = \frac{n + 1}{2}$

The deviations of numbers from the mean are as shown below

$1 - \frac{n + 1}{2}, 2 - \frac{n + 1}{2}, \dots \dots \dots \dots (n - 1) - \frac{n - 1}{2}, n - \frac{n + 1}{2}$

Or, $\frac{2 - (n + 1)}{2}, \frac{4 - (n + 1)}{2} \dots \dots \dots . \frac{2 (n - 1) - (n + 1)}{2}, \frac{2 n - (n + 1)}{2}$

$Or, \frac{1 - n}{2}, \frac{3 - n}{2}, \frac{5 - n}{2} \dots \dots . \frac{n - 3}{2}, \frac{n - 1}{2}$

$\frac{- (n - 1)}{2}, \frac{- (n - 3)}{2}, \frac{- (n - 5)}{2}, \dots \dots, \frac{n - 5}{2}, \frac{n - 3}{2}, \frac{n - 1}{2}$

$So the absolute values of deviation from the mean is | x_{i} - \overset{―}{x} | = \frac{n - 1}{2}, \frac{n - 3}{2}, \frac{n - 5}{2}, \dots \dots . \frac{n - 5}{2}, \frac{n - 3}{2}, \frac{n - 1}{2}$

$Σ | x_{i} - \overset{―}{x} | = 2 (1 + 2 + 3 + \dots \dots . + \frac{n - 5}{2} + \frac{n - 3}{2} + \frac{n - 1}{2})$

$That is 2 times the sum of \frac{n - 1}{2} terms$

$So, it can be written as Σ | x_{i} - \overset{―}{x} | = 2 (\frac{\frac{n - 1}{2} (\frac{n - 1}{2} + 1)}{2}) = n (\frac{n - 1}{2}) (\frac{n + 1}{2}) = (\frac{n^{2} - 1}{4})$

$Therefore, mean deviation about the mean is \frac{Σ | x_{i} - \overset{―}{x} |}{n} = \frac{(\frac{n^{2} - 1}{4})}{n} = (\frac{n^{2} - 1}{4 n})$

Question:4

Calculate the mean deviation about the mean of the set of the first n natural numbers when n is an even number.

Answer:

A set of first n natural numbers when n is an even number is given

We have to find the mean deviation from the mean

We know that fthe irst natural numbers are 1, 2, 3 ……. , n.

It is given that n is an even number.

So, the mean is

$\overset{―}{x} = \frac{1 + 2 + 3 + \dots . . + n}{n} = \frac{\frac{n (n + 1)}{2}}{n} = \frac{n + 1}{2}$

The deviations of numbers from the mean are as shown below:

$1 - \frac{n + 1}{2}, 2 - \frac{n + 1}{2}, \dots \dots \dots \dots (n - 1) - \frac{n - 1}{2}, n - \frac{n + 1}{2}$

$O r, \frac{2 - (n + 1)}{2}, \frac{4 - (n + 1)}{2} \dots \dots \dots . \frac{2 (n - 1) - (n + 1)}{2}, \frac{2 n - (n + 1)}{2}$
$Or, \frac{1 - n}{2}, \frac{3 - n}{2}, \frac{5 - n}{2} \dots \dots . \frac{n - 3}{2}, \frac{n - 1}{2}$

= $\frac{- (n - 1)}{2}, \frac{- (n - 3)}{2}, \frac{- (n - 5)}{2}, \dots \dots, \frac{n - 5}{2}, \frac{n - 3}{2}, \frac{n - 1}{2}$

$The sum of absolute values of deviations from the mean, is Σ | x_{i} - \overset{―}{x} | = \frac{n - 1}{2} + \frac{n - 3}{2} + \frac{n - 5}{2} + \dots \dots \frac{n - 5}{2} + \frac{n - 3}{2} + \frac{n - 1}{2} Σ | x_{i} - \overset{―}{x} | = (\frac{1}{2} + \frac{3}{2} + \dots \dots . + \frac{n - 5}{2} + \frac{n - 3}{2} + \frac{n - 1}{2}) (\frac{n}{2})$

$We know that sum of first n natural numbers = (\frac{n^{2} + n}{2})$

$Therefore, mean deviation about the mean is \frac{Σ | x_{i} - \overset{―}{x} |}{n} = \frac{(\frac{1}{2} + \frac{3}{2} + \dots . . + \frac{n - 1}{2}) (\frac{n}{2})}{n} = \frac{{(\frac{n}{2})}^{2}}{n} = \frac{n^{2}}{4 n} = \frac{n}{4}$

Question:5

Find the standard deviation of the first n natural numbers.

Answer:

$Set of first n natural numbers is given.$

$We have to find the standard deviation.$

$We can write in the table the first n natural numbers as$
$\begin{array}{cllllllll} x_{i} & 1 & 2 & 3 & 4 & 5 & \dots . & \dots . & n \\ x_{i}^{2} & 1 & 4 & 9 & 16 & 25 & \dots . & \dots \dots & n^{2} \end{array}$

$The sums are Σ x_{i} = 1 + 2 + 3 + \dots \dots + n = \frac{n (n + 1)}{2}$

$Σ x_{i}^{2} = 1^{2} + 2^{2} + 3^{2} + \dots \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

Therefore, the standard deviation can be written as, $σ = \sqrt{\frac{Σ x_{i}^{2}}{n} - {(\frac{Σ x_{i}}{n})}^{2}}$

$\Rightarrow σ = \sqrt{\frac{\frac{n (n + 1) (2 n + 1)}{6}}{n} - {(\frac{\frac{n (n + 1)}{2}}{n})}^{2}} = \sqrt{\frac{2 n^{2} + n + 2 n + 1}{6} - \frac{n^{2} + 2 n + 1}{4}} = \sqrt{\frac{4 n^{2} + 6 n + 2 - 3 n^{2} - 6 n - 3}{12}} = \sqrt{\frac{n^{2} - 1}{12}}$

Question:6

The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds.
Further, another set of 15 observations x₁, x₂, ..., x₁₅, also in seconds, is now available and we have $\sum_{i = 1}^{15} x_{i} = 279$ and $\sum_{i = 1}^{15} x_{i} = 5524$ .Calculate the standard derivation based on all 40 observations.

Answer:

Number of observations=25, mean = 18.2 seconds, standard deviation = 3.25 seconds.

$There are another set of 15 observations x_{1}, x_{2}, \dots . ., x_{15}$

$Also in seconds is \sum_{i = 1}^{15} x_{i} = 279 a n d \sum_{i = 1}^{15} x_{i}^{2} = 5524$

$To find that the standard deviation based on all 40 observations$

$As per the given criteria in the first set Number of observations, n_{1} = 25$

Mean=18.2

$And standard deviation σ_{1} = 3.25$

$And in second set number of observations n_{2} = 15$

$For the first set we have \overset{―}{x_{1}} = 18.2 = \frac{Σ x_{i}}{25}$

$\Rightarrow Σ x_{i} = 25 \times 18.2 = 455$
$Therefore the standard deviation becomes σ_{1}^{2} = \frac{Σ x_{i}^{2}}{25} - {(18.2)}^{2}$

$Substituting the values, we get {(3.25)}^{2} = \frac{Σ x_{i}^{2}}{25} - 331.24 10.5625 + 331.24 = \frac{Σ x_{i}^{2}}{25} = 341.8025 Σ x_{i}^{2} = 25 \times 341.8025 = 8545.06$

$For the combined standard deviation of the 40 observation, n=40 and Σ x_{i}^{2} = 8545.06 + 5524 = 14069.69 Σ x_{i} = 455 + 279 = 734$

$Therefore the standard deviation can be written as, σ = \sqrt{\frac{Σ x_{i}^{2}}{n} - {(\frac{Σ x_{i}}{n})}^{2}}$
$Substituting the values, we get Therefore the standard deviation can be written as σ = \sqrt{\frac{14069.69}{40} - {(\frac{734}{40})}^{2}} = \sqrt{351.7265 - 336.7225} = \sqrt{15.004} = 3.87$

Question:7

The mean and standard deviation of a set of $n_{1}$ observations are $\bar{X_{1}}$ and $s_{1}$ , respectively while the mean and standard deviation of another set of $n_{2}$ observations are $\bar{X_{2}}$ and $s_{2}$ , respectively. Show that the standard deviation of the combined set of ( $n_{1} + n_{2}$ ) observations is given by
$SD = \sqrt{\frac{n_{1} {(s_{1})}^{2} + n_{2} {(s_{2})}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} {({\overset{―}{x}}_{1} - {\overset{―}{x}}_{2})}^{2}}{{(n_{1} + n_{2})}^{2}}}$

Answer:

$It is given that the mean and Standard deviation of a set of n_{1} observations are \overset{―}{x_{1}} a n d σ_{1} r e s p e c t i v e l y$

$While the mean and standard deviation of naother set of n_{2} observations are \overset{―}{x_{2}} a n d σ_{2} r e s p e c t i v e l y \]$

$\overset{―}{x_{1}} = \frac{1}{n_{1}} \sum_{i = 1}^{n_{1}} x_{i} \overset{―}{x_{2}} = \frac{1}{n_{2}} \sum_{j = 1}^{n_{2}} y_{j} \overset{―}{x} = \frac{1}{(n_{1} + n_{2})} [\sum_{i = 1}^{n_{1}} x_{i} + \sum_{j = 1}^{n_{2}} y_{j}] = \frac{n_{1} \overset{―}{x_{1}} + n_{2} \overset{―}{x_{2}}}{n_{1} + n_{2}} σ_{1}^{2} = \frac{1}{n_{1}} \sum_{i = 1}^{n_{1}} {(x_{i} - \overset{―}{x})}^{2}$

$σ_{2}^{2} = \frac{1}{n_{2}} \sum_{j = 1}^{n_{2}} {(y_{j} - \overset{―}{x})}^{2} σ^{2} = σ_{1}^{2} + σ_{2}^{2} = \frac{1}{n_{1} + n_{2}} [\sum_{i = 1}^{n_{1}} {(x_{i} - \overset{―}{x})}^{2} + \frac{1}{n_{2}} \sum_{j = 1}^{n_{2}} {(y_{j} - \overset{―}{x})}^{2}] \sum_{i = 1}^{n_{1}} {(x_{i} - \overset{―}{x})}^{2} = \sum_{i = 1}^{n_{1}} {(x_{i} - \overset{―}{x_{j}} + \overset{―}{x_{j}} - \overset{―}{x})}^{2} = \sum_{i = 1}^{n_{1}} {(x_{i} - \overset{―}{x_{j}})}^{2} + n_{1} {(\overset{―}{x_{j}} - \overset{―}{x})}^{2} + 2 (\overset{―}{x_{j}} - \overset{―}{x}) \sum_{i = 1}^{n_{1}} (x_{i} - \overset{―}{x_{j}}) \sum_{i = 1}^{n_{1}} (x_{i} - \overset{―}{x_{j}}) = 0$

$\sum_{i = 1}^{n_{1}} {(x_{i} - \overset{―}{x})}^{2} = n_{1} s_{1}^{2} + n_{1} {(\overset{―}{x_{1}} - \overset{―}{x})}^{2} d_{1} = \overset{―}{x_{1}} - \overset{―}{x} = \overset{―}{x_{1}} - \frac{n_{1} \overset{―}{x_{1}} + n_{2} \overset{―}{x_{2}}}{n_{1} + n_{2}} = \frac{n_{2} (\overset{―}{x_{1}} - \overset{―}{x_{2}})}{n_{1} + n_{2}} \sum_{i = 1}^{n_{1}} {(x_{i} - \overset{―}{x})}^{2} = n_{1} s_{1}^{2} + \frac{n_{1} n_{2}^{2} {(\overset{―}{x_{1}} - \overset{―}{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}} \sum_{j = 1}^{n_{2}} {(y_{j} - \overset{―}{x})}^{2} = \sum_{j = 1}^{n_{2}} {(y_{j} - \overset{―}{x_{i}} + \overset{―}{x_{i}} - \overset{―}{x})}^{2}$
$= \sum_{j = 1}^{n_{2}} {(y_{j} - \overset{―}{x_{i}})}^{2} + n_{2} {(\overset{―}{x_{i}} - \overset{―}{x})}^{2} + 2 (\overset{―}{x_{i}} - \overset{―}{x}) \sum_{j = 1}^{n_{2}} (y_{j} - \overset{―}{x_{i}}) \sum_{j = 1}^{n_{2}} (y_{j} - \overset{―}{x_{i}}) = 0 \sum_{j = 1}^{n_{2}} {(y_{j} - \overset{―}{x})}^{2} = n_{2} s_{2}^{2} + n_{2} {(\overset{―}{x_{2}} - \overset{―}{x})}^{2} d_{2} = \overset{―}{x_{2}} - \overset{―}{x} = \overset{―}{x_{2}} - \frac{n_{1} \overset{―}{x_{1}} + n_{2} \overset{―}{x_{2}}}{n_{1} + n_{2}} = \frac{n_{1} (\overset{―}{x_{2}} - \overset{―}{x_{1}})}{n_{1} + n_{2}}$
$\sum_{j = 1}^{n_{2}} {(x_{i} - \overset{―}{x})}^{2} = n_{2} s_{2}^{2} + \frac{n_{1}^{2} n_{2} {(\overset{―}{x_{2}} - \overset{―}{x_{1}})}^{2}}{{(n_{1} + n_{2})}^{2}} σ^{2} = σ_{1}^{2} + σ_{2}^{2} = \frac{1}{n_{1} + n_{2}} [\sum_{i = 1}^{n_{1}} {(x_{i} - \overset{―}{x})}^{2} + \sum_{j = 1}^{n_{2}} {(y_{j} - \overset{―}{x})}^{2}] = \frac{1}{n_{1} + n_{2}} [n_{1} s_{1}^{2} + \frac{n_{1} n_{2}^{2} {(\overset{―}{x_{1}} - \overset{―}{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}} + n_{2} s_{2}^{2} + \frac{n_{1}^{2} n_{2} {(\overset{―}{x_{2}} - \overset{―}{x_{1}})}^{2}}{{(n_{1} + n_{2})}^{2}}] = \frac{1}{n_{1} + n_{2}} [n_{1} s_{1}^{2} + n_{2} s_{2}^{2} + \frac{n_{1} n_{2} {(\overset{―}{x_{1}} - \overset{―}{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}} (n_{1} + n_{2})]$

$= \frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} {(\overset{―}{x_{1}} - \overset{―}{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}} σ = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} {(\overset{―}{x_{1}} - \overset{―}{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}}}$

Question:8

Two sets each of 20 observations have the same standard deviation of 5. The first set has a mean of 17 and the second a mean of 22. Determine the standard deviation of the set obtained by combining the given two sets.

Answer:

$Given that two sets, each of 20 observations, have the same standard deviation 5.$

$The first set has a mean of 17 and the second a mean of 22.$

$We have to show that the standard deviation of the set obtained by combining the given two sets.$

$As per given criteria, for first set Number of observations, n_{1} = 20$

$Standard deviation s_{1} = 5 And mean \overset{―}{x_{1}} = 17$

$For second set Number of observations, n_{2} = 20 Standard deviation s_{2} = 5 and mean \overset{―}{x_{2}} = 22$

$We know the standard deviation for combined two series is σ = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + (n_{1} n_{2}) \frac{{(\overset{―}{x_{1}} - \overset{―}{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}}}$

$Substituting the corresponding values we get σ = \sqrt{\frac{20 {(5)}^{2} + 20 {(5)}^{2}}{20 + 20} + \frac{(20 * 20) {(17 - 22)}^{2}}{{(20 + 20)}^{2}}} = \sqrt{\frac{1000}{40} + \frac{10000}{1600}} = \sqrt{25 + \frac{25}{4}} = \sqrt{\frac{100 + 25}{4}} = \sqrt{\frac{125}{4}} = 5.59$

Question:9

The frequency distribution:
$\begin{array}{lllllll} x & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ f & 2 & 1 & 1 & 1 & 1 & 1 \end{array}$
where A is a positive integer, has a variance of 160. Determine the value of A.

Answer:

A frequency distribution table is given where variance =160. We have to find the value of A, where A is a positive number. Let us make a table from the given data and fill out the other columns after the calculation
$\begin{array}{lllc} Size (x_{i}) & Frequency (f_{i}) & f_{i} x_{i} & f_{i} x_{i}^{2} \\ A & 2 & 2 A & 2 A^{2} \\ 2 A & 1 & 2 A & 4 A^{2} \\ 3 A & 1 & 3 A & 9 A^{2} \\ 4 A & 1 & 4 A & 16 A^{2} \\ 5 A & 1 & 5 A & 25 A^{2} \\ 6 A & 1 & 6 A & 36 A^{2} \\ Total & N = 7 & 22 A & 92 A^{2} \end{array}$

And we know that variance is $σ^{2} = \frac{Σ f_{i} x_{i}^{2}}{n} - {(\frac{Σ f_{i} x_{i}}{n})}^{2}$

$Substituting the values from the above table and also given that variance = 160 w e g e t 160 = \frac{92 A^{2}}{7} - {(\frac{22 A}{7})}^{2} = (\frac{92 A^{2}}{7} - \frac{484 A^{2}}{49}) = \frac{7 * 92 A^{2} - 484 A^{2}}{49} = \frac{160 A^{2}}{49}$

$160 = \frac{160 A^{2}}{49}$

$\Rightarrow A^{2} = 49$

$\Rightarrow A = 7$

$Hence, the value of A is 7$

Question:10

For the frequency distribution:
$\begin{array}{lllllll} x & 2 & 3 & 4 & 5 & 6 & 7 \\ f & 4 & 9 & 16 & 14 & 11 & 6 \end{array}$
Find the standard distribution.

Answer:

A frequency distribution table is given, and we have to find the standard deviation

Let us make a table from the given data and fill out the other columns after calculation

$\begin{array}{llll} Size (x_{i}) & Frequency (f_{i}) & f_{i} x_{i} & f_{i} x_{i}^{2} \\ 2 & 4 & 8 & 16 \\ 3 & 9 & 27 & 81 \\ 4 & 16 & 64 & 256 \\ 5 & 14 & 70 & 350 \\ 6 & 11 & 66 & 396 \\ 7 & 6 & 42 & 294 \\ Total & N = 60 & 277 & 1393 \end{array}$

$\begin{aligned} And we know that standard deviation is \\ σ = \sqrt{\frac{Σ f_{i} x_{i}^{2}}{n} - {(\frac{Σ f_{i} x_{i}}{n})}^{2}} \end{aligned}$

Substituting the value from the above table

$σ = \sqrt{\frac{1393}{60} - {(\frac{277}{60})}^{2}} = \sqrt{23.23 - 21.34} = 1.37$

Question:11

There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
$\begin{array}{lllllll} Marks & 0 & 1 & 2 & 3 & 4 & 5 \\ Frequency & x - 2 & x & x^{2} & (x + 1)^{2} & 2 x & x + 1 \end{array}$
where x is a positive integer. Determine the mean and standard deviation of the marks.

Answer:

It is given that there are 60 students in a class. The frequency distribution of the marks obtained by the students in a test is also given.

We have to find the mean and standard deviation of the marks.

It is given that there are 60 students in the class, so

$Σ f_{i} = 60 (x - 2) + x + x^{2} + {(x + 1)}^{2} + 2 x + x + 1 = 60 5 x - 1 + x^{2} + x^{2} + 2 x + 1 = 60 2 x^{2} + 7 x = 60$

$On factorising 2 x^{2} + 7 x - 60 = 0$

$we get (2 x + 15) (x - 4) = 0 2 x = - 15 o r x = 4$

$Given that x is a positive number, so x can take 4 as the only value .$

$And let assumed mean, a = 3$

$Now put x=4 and a=3 in the frequency distribution table and other columns after calculations, we get$

$\begin{array}{lllll} Marks (x_{i}) & Frequency (f_{i}) & d_{i} = x_{i} - a_{i} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ 0 & x - 2 = 4 - 2 = 2 & - 3 & - 6 & 18 \\ 1 & X = 4 & - 2 & - 8 & 16 \\ 2 & x^{2} = 16 & - 1 & - 16 & 16 \\ 3 & (x + 1)^{2} & 0 & 0 & 0 \\ = (4 + 1)^{2} = 25 \\ 4 & 2 x = 2 \times 4 = 8 & 1 & 4 & 8 \\ 5 & X + 1 = 4 + 1 = 5 & 2 & 10 & 20 \\ Total & N = 60 & 277 & = - 12 & 78 \end{array}$

And we know that standard deviation is $σ = \sqrt{\frac{Σ f_{i} d_{i}^{2}}{n} - {(\frac{Σ f_{i} d_{i}}{n})}^{2}}$

$Substituting the values from the above table σ = \sqrt{\frac{78}{60} - {(\frac{- 12}{60})}^{2}} = \sqrt{1.3 - {(0.2)}^{2}} = \sqrt{1.3 - 0.04} = 1.12$

$Hence, the standard deviation is 1.12$

$Now mean is \overset{―}{x} = A + \frac{Σ f_{i} d_{i}}{N} = 3 + (- \frac{12}{60}) = 3 - \frac{1}{5} = \frac{14}{5} = 2.8$

$Hence, the mean and standard deviation of the marks are 2.8 and 1.12, respectively$

Question:12

The mean life of a sample of 60 bulbs was 650 hours, and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and a standard deviation of 7 hours. Find the overall standard deviation.

Answer:

The mean life of a sample of 60 bulbs was 650 hours, and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and a standard deviation of 7 hours. We have to find the overall standard deviation.

As per the given criteria. In the first set of samples number of sample bulbs, $n_{1} = 60$

$Standard deviation s_{1} = 8 hrs Mean life, \overset{―}{x_{1}} = 650$

$And in the second set of samples, number of sample bulbs n_{2} = 80 Standard deviation s_{2} = 7 h r s$

$Mean life \overset{―}{x_{2}} = 660$

$Substituting the corresponding values we get σ = \sqrt{\frac{60 {(8)}^{2} + 80 {(7)}^{2}}{60 + 80} + \frac{(60 * 80) {(650 - 660)}^{2}}{{(20 + 80)}^{2}}} = \sqrt{\frac{388}{7} + \frac{1200}{49}} = = \sqrt{\frac{3916}{49}} = 8.9$

Question:13

Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all items and the sum of the squares of the items.

Answer:

The mean and standard deviation of 100 items are 50 and 4, respectively.
We have to find the sum of all items and the sum of the squares of the items.
As per the question, the number of items n =100
$Mean of the given items, \overset{―}{x} = 50$

$But we know, \overset{―}{x} = \frac{Σ x_{i}}{n}$

$Substituting the corresponding values we get 50 = \frac{Σ x_{i}}{100} Σ x_{i} = 50 * 100 = 5000$

$Hence, the sum of all the 100 items = 5000$

$Also given that the standard deviation of the 100 items is 4 σ = 4$

$But we know that \sqrt{\frac{Σ x_{i}^{2}}{n} - {(\frac{Σ x_{i}}{n})}^{2}}$

$Substituting the corresponding values, we get 4 = \sqrt{\frac{Σ x_{i}^{2}}{100} - {(\frac{5000}{100})}^{2}}$

$Now taking square on both sides, we get 4^{2} = \frac{Σ x_{i}^{2}}{100} - {(50)}^{2}$
$16 = \frac{Σ x_{i}^{2}}{100} - 2500$

$\Rightarrow 16 + 2500 = \frac{Σ x_{i}^{2}}{100}$

$\frac{Σ x_{i}^{2}}{100} = 2516$

$Σ x_{i}^{2} = 2516 \times 100 = 251600$

Question:14

If for a distribution $$ \sum $(x$ - $5) = 3,$ \sum $(x$ - $5)^{2} = 43$ and the total number of item is 18, find the mean and standard deviation.

Answer:

$It is given that for a distribution Σ (x - 5) = 3$

$Σ {(x - 5)}^{2} = 43 and the total number of item is 18$

$We have to find the mean and standard deviation As per given criteria, Number of items, n=18$

$Mean can be written as \overset{―}{x} = A + \frac{Σ (x - 5)}{n}$

$Here assumed mean is 5, so substituting the corresponding values in the above equation we get \overset{―}{x} = 5 + \frac{3}{18} = \frac{93}{18} = 5.17$

$And we know that the standard deviation can be written as, σ = \sqrt{\frac{Σ {(x - 5)}^{2}}{n} - {(\frac{Σ (x - 5)}{n})}^{2}}$

Substituting the corresponding values, we get

$σ = \sqrt{\frac{43}{18} - {(\frac{3}{18})}^{2}} = \sqrt{2.39 - (0.166)^{2}} = \sqrt{2.39 - 0.027} = \sqrt{2.363} = 1.54$

Thus, the mean and standard deviation of the given items are 5.17 and 1.54, respectively.

Question:15

Find the mean and variance of the frequency distribution given below:
$\begin{array}{lllll} x & 1 \leq x < 3 & 3 \leq x < 5 & 5 \leq x < 7 & 7 \leq x < 10 \\ f & 6 & 4 & 5 & 1 \end{array}$

Answer:

Frequency distribution is given. We have to find the mean and variance

Converting the ranges of x to groups, the given table can be rewritten as

$\begin{array}{lllll} X (Class) & Size (x_{i}) & Frequency (f_{i}) & f_{i} x_{i} & f_{i} x_{i}^{2} \\ 1 - 3 & 2 & 6 & 12 & 24 \\ 3 - 5 & 4 & 4 & 16 & 64 \\ 5 - 7 & 6 & 5 & 30 & 180 \\ 7 - 10 & 8.5 & 1 & 8.5 & 72.25 \\ Total & N = 16 & = 66.5 & 340.25 \end{array}$

And, we know that variance is $σ^{2} = \frac{Σ f_{i} x_{i}^{2}}{n} - {(\frac{Σ f_{i} x_{i}}{n})}^{2}$

$Substituting values from above table, we get σ^{2} = \frac{340.25}{16} - {(\frac{66.5}{16})}^{2}$

$σ^{2} = 21.265 - {(4.16)}^{2} = 21.265 - 17.305 = 3.96$

$mean \overset{―}{X} = \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{66.5}{16} = 4.16$

$Hence, the mean and variance of the given frequency distribution are 4.16 and 3.96, respectively$

Question:16

Calculate the mean deviation about the mean for the following frequency distribution:
$\begin{array}{llllll} Class interval & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 \\ Frequency & 4 & 6 & 8 & 5 & 2 \end{array}$

Answer:

A frequency distribution table is given, and we have to find the mean deviation from the mean
Let us make a table from the given data and fill out the other columns after calculation
$\begin{array}{llll} \begin{matrix} Class \\ Interval \end{matrix} & \begin{matrix} Mid value \\ (x_{i}) \end{matrix} & \begin{matrix} Frequency \\ (f_{i}) \end{matrix} & f_{i} x_{i} \\ 0 - 4 & 2 & 4 & 8 \\ 4 - 8 & 6 & 6 & 36 \\ 8 - 12 & 10 & 8 & 80 \\ 12 - 16 & 14 & 5 & 70 \\ 16 - 20 & 18 & 2 & 36 \\ Total & N = 25 & = 230 \end{array}$
$m e a n \overset{―}{X} = \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{230}{25} = 9.2$

$The above column can be rewritten as$

$\begin{array}{llllll} \begin{matrix} Class \\ Interval \end{matrix} & \begin{matrix} Mid value \\ (x_{i}) \end{matrix} & \begin{matrix} Frequency \\ (f_{i}) \end{matrix} & f_{i} x_{i} & \begin{matrix} d_{i} =∣ x_{i} - \\ \bar{x} ∣ \end{matrix} & f_{i} d_{i} \\ 0 - 4 & 2 & 4 & 8 & 7.2 & 28.8 \\ 4 - 8 & 6 & 6 & 36 & 3.2 & 19.2 \\ 8 - 12 & 10 & 8 & 80 & 0.8 & 6.4 \\ 12 - 16 & 14 & 5 & 70 & 1.8 & 24.4 \\ 16 - 20 & 18 & 2 & 36 & 8.8 & 17.6 \\ Total & N = 25 & = 230 & = 96.0 \end{array}$
$m e a n D e v i a t i o n = \frac{Σ f_{i} d_{i}}{Σ f_{i}} = \frac{96}{25} = 3.84$

Question:17

Calculate the mean deviation from the median of the following data:
$\begin{array}{llllll} Class interval & 0 - 6 & 6 - 12 & 12 - 18 & 18 - 24 & 24 - 30 \\ Frequency & 4 & 5 & 3 & 6 & 2 \end{array}$

Answer:

A frequency distribution table is given, and we have to find the mean deviation from the median

Let us make a table from the given data and fill out the other columns after calculation

$\begin{array}{llll} \begin{matrix} Class \\ Interval \end{matrix} & \begin{matrix} Mid value \\ (x_{i}) \end{matrix} & \begin{matrix} Frequency \\ (f_{i}) \end{matrix} & \begin{matrix} Cumulative \\ frequency \\ (c_{f} f) \end{matrix} \\ 0 - 6 & 3 & 4 & 4 \\ 6 - 12 & 9 & 5 & 9 \\ 12 - 18 & 15 & 3 & 12 \\ 18 - 24 & 21 & 6 & 18 \\ 24 - 30 & 27 & 2 & 20 \\ Total & N = 20 \end{array}$

Now here N=20, which is even

Here $Median Class = \frac{N}{2} = 10^{t h} t e r m$

$This observation lies in the class interval 12-18, so the median can be written as M = l + \frac{\frac{N}{2} - c f}{f} \times h$

$Here l=12, cf=9, f=3, h=6 and N=20$

$Substituting these values, the above equation becomes, M = 12 + \frac{\frac{20}{2} - 9}{3} \times 6$

$\Rightarrow M = 12 + \frac{10 - 9}{3} \times 6$

$= 12 + \frac{1 \times 6}{3}$

$\Rightarrow M = 12 + 2 = 14$

$\begin{array}{lllll} \begin{matrix} Class \\ Interval \end{matrix} & \begin{matrix} Mid value \\ (x_{i}) \end{matrix} & \begin{matrix} Frequency \\ (f_{i}) \end{matrix} & d_{i} = | x_{i} - M | & f_{i} d_{i} \\ 0 - 6 & 3 & 4 & 11 & 44 \\ 6 - 12 & 9 & 5 & 5 & 25 \\ 12 - 18 & 15 & 3 & 1 & 3 \\ 18 - 24 & 21 & 6 & 7 & 42 \\ 24 - 30 & 27 & 2 & 13 & 26 \\ Total & N = 20 & = 140 \end{array}$

$m e a n D e v i a t i o n = \frac{Σ f_{i} d_{i}}{Σ f_{i}} = \frac{140}{20} = 7$

Question:18

Determine the mean and standard deviation for the following distribution:
$\begin{array}{llllllllllllllll} Marks & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ Frequency & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \end{array}$

Answer:

A frequency distribution table is given, and we have to find the mean and standard deviation
Let us make a table from the given data and fill out the other columns after calculation

$\begin{array}{lll} Marks (x_{i}) & \begin{matrix} Frequency \\ (f_{i}) \end{matrix} & f_{i} x_{i} \\ 2 & 1 & 2 \\ 3 & 6 & 18 \\ 4 & 6 & 24 \\ 5 & 8 & 40 \\ 6 & 8 & 48 \\ 7 & 2 & 14 \\ 8 & 2 & 16 \\ 9 & 3 & 27 \\ 10 & 0 & 0 \\ 11 & 2 & 22 \\ 12 & 1 & 12 \\ 13 & 0 & 0 \\ 14 & 0 & 0 \\ 15 & 0 & 0 \\ 16 & 1 & 16 \\ Total & N = 40 & = 239 \end{array}$
$m e a n \overset{―}{X} = \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{239}{40} = 5.975 \approx 6$
$\begin{array}{llllll} Marks (x_{i}) & \begin{matrix} Frequency \\ (f_{i}) \end{matrix} & f_{i} x_{i} & d_{i} = x_{i} - \bar{x} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ 2 & 1 & 2 & - 4 & - 4 & 16 \\ 3 & 6 & 18 & - 3 & - 18 & 54 \\ 4 & 6 & 24 & - 2 & - 12 & 24 \\ 5 & 8 & 40 & - 1 & - 8 & 8 \\ 6 & 8 & 48 & 0 & 0 & 0 \\ 7 & 2 & 14 & 1 & 2 & 2 \\ 8 & 2 & 16 & 2 & 4 & 8 \\ 9 & 3 & 27 & 3 & 9 & 27 \\ 10 & 0 & 0 & 4 & 0 & 0 \\ 11 & 2 & 22 & 5 & 10 & 50 \\ 12 & 1 & 12 & 6 & 6 & 36 \\ 13 & 0 & 0 & 7 & 0 & 0 \\ 14 & 0 & 0 & 8 & 0 & 0 \\ 15 & 0 & 0 & 9 & 0 & 0 \\ 16 & 1 & 16 & 10 & 10 & 100 \\ Total & N = 40 & = 239 & = - 1 & = 325 \end{array}$
And we know that standard deviation is
$σ = \sqrt{\frac{Σ f_{i} d_{i}^{2}}{n} - {(\frac{Σ f_{i} d_{i}}{n})}^{2}}$

$= \sqrt{\frac{325}{40} - {(\frac{- 1}{40})}^{2}}$

$= \sqrt{8.125 - {(0.025)}^{2}} = 2.8721$

Question:19

The weights of coffee in 70 jars are shown in the following table:
$\begin{array}{ll} Weight (in grams) & Frequency \\ 200 - 201 & 13 \\ 201 - 202 & 27 \\ 202 - 203 & 18 \\ 203 - 204 & 10 \\ 204 - 205 & 1 \\ 205 - 206 & 1 \end{array}$
Determine the variance and standard deviation of the above distribution.

Answer:

A frequency distribution table for the weights of coffee in 70 jars is given, and we have to find the variance and standard deviation of the distribution.
Let us make a table from the given data and fill out the other columns after calculation
$\begin{array}{llll} \begin{matrix} Weight (in \\ grams) \end{matrix} & \begin{matrix} Mid-value \\ (x_{i}) \end{matrix} & \begin{matrix} Frequency \\ (f_{i}) \end{matrix} & f_{i} x_{i} \\ 200 - 201 & 200.5 & 13 & 2606.5 \\ 201 - 202 & 201.5 & 27 & 5440.5 \\ 202 - 203 & 202.5 & 18 & 3645 \\ 203 - 204 & 203.5 & 10 & 2035 \\ 204 - 205 & 204.5 & 1 & 204.5 \\ 205 - 206 & 205.5 & 1 & 205.5 \\ total & N = 70 & = 14137 \end{array}$
$m e a n \overset{―}{X} = \frac{Σ f_{i} x_{i}}{N} = \frac{14137}{70} = 201.9$
$\begin{array}{llllll} \begin{matrix} Weight (in \\ grams) \end{matrix} & \begin{matrix} Mid-value \\ (x_{i}) \end{matrix} & \begin{matrix} Frequency \\ (f_{i}) \end{matrix} & d_{i} = x_{i} - \bar{x} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ 200 - 201 & 200.5 & 13 & - 1.4 & - 18.2 & 25.48 \\ 201 - 202 & 201.5 & 27 & - 0.4 & - 10.8 & 4.32 \\ 202 - 203 & 202.5 & 18 & 0.6 & 10.8 & 6.48 \\ 203 - 204 & 203.5 & 10 & 1.6 & 16 & 25.6 \\ 204 - 205 & 204.5 & 1 & 2.6 & 2.6 & 6.76 \\ 205 - 206 & 205.5 & 1 & 3.6 & 3.6 & 12.96 \\ total & N = 70 & = 4 & = 81.6 \end{array}$
And we know that standard deviation is
$σ = \sqrt{\frac{Σ f_{i} d_{i}^{2}}{n} - {(\frac{Σ f_{i} d_{i}}{n})}^{2}}$

$= \sqrt{\frac{81.6}{70} - {(\frac{4}{70})}^{2}}$

$= \sqrt{1.17 - {(0.057)}^{2}}$

$= \sqrt{1.17} = 1.08$
Variance = 1.17

Question:20

Determine the mean and standard deviation of the first n terms of an A.P. whose first term is a and common difference is d.

Answer:

The first n terms of an A.P are given whose first term is a and the common difference is d. We have to find the mean and standard deviation.

The given AP in tabular form is as shown below,

$\begin{array}{lll} | c | x_{i} & | c | d_{i} = x_{i} - a & d_{i}^{2} \\ a & 0 & 0 \\ a + d & d & d^{2} \\ a + 2 d & 2 d & 4 d^{2} \\ a + 3 d & 3 d & 9 d^{2} \\ - & - & - \\ a + (n - 1) d & (n - 1) d & (n - 1)^{2} d^{2} \end{array}$
$Here we have assumed a as mean Given that AP has n terms. And we know the sum of all terms of AP can be written as$

$\sum x_{i} = \frac{n}{2} [2 a + (n - 1) d] \overset{―}{x}$

$\Rightarrow \frac{\sum x_{i}}{n} = \frac{2 a + (n - 1) d}{2} = a + \frac{(n - 1)}{2} d$

$\sum d_{i} = \sum (x_{i} - a)$

$= d [1 + 2 + 3 + \dots + (n - 1)] = d (\frac{n (n - 1)}{2})$

$\sum d_{i}^{2} = \sum {(x_{i} - a)}^{2}$

$= d^{2} [1^{2} + 2^{2} + 3^{2} + \dots + {(n - 1)}^{2}] = d^{2} (\frac{n (n - 1) (2 n - 1)}{6})$

$σ = \sqrt{\frac{\sum {(x_{i} - a)}^{2}}{n} - {(\frac{\sum (x_{i} - a)}{n})}^{2}}$

$= \sqrt{\frac{d^{2} (\frac{n (n - 1) (2 n - 1)}{6})}{n} - \frac{d^{2} n^{2} {(n - 1)}^{2}}{4 n^{2}}}$

$= \sqrt{\frac{d^{2} (n - 1) (2 n - 1)}{6} - \frac{d^{2} {(n - 1)}^{2}}{4}}$

$= \sqrt{\frac{d^{2} (n - 1)}{2} (\frac{2 n - 1}{3} - \frac{n - 1}{2})}$

$= \sqrt{\frac{d^{2} (n - 1)}{2} (\frac{4 n - 2 - 3 n + 3}{6})} = \sqrt{\frac{d^{2} (n - 1)}{2} (\frac{n + 1}{6})} = d \sqrt{\frac{n^{2} - 1}{12}}$

Question:21

Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.
$\begin{array}{lllllllllll} Ravi & 25 & 50 & 45 & 30 & 70 & 42 & 36 & 48 & 35 & 60 \\ Hashina & 10 & 70 & 50 & 20 & 95 & 55 & 42 & 60 & 48 & 80 \end{array}$
Who is more intelligent and who is more consistent?

Answer:

The marks obtained, out of 100, by 2 students Ravi and Hashina in 10 tests are given
We have to find who is more intelligent and who is more consistent
The marks of Ravi taken separately as follows
$\begin{array}{lll} M a r k s (x_{i}) & d_{i} = x_{i} - 45 & d_{i}^{2} \\ 25 & - 20 & 400 \\ 50 & 5 & 25 \\ 45 & 0 & 0 \\ 30 & - 15 & 225 \\ 70 & 25 & 625 \\ 42 & - 3 & 9 \\ 36 & - 9 & 81 \\ 48 & 3 & 9 \\ 35 & - 10 & 100 \\ 60 & 15 & 225 \\ TOTAL = 441 & = - 9 & = 1699 \end{array}$
Here we have assumed 45 as mean
And we know that the standard deviation can be written as,
$σ = \sqrt{\frac{Σ {(x - a)}^{2}}{n} - {(\frac{Σ (x - a)}{n})}^{2}}$

$\Rightarrow σ = \sqrt{\frac{1699}{10} - {(- \frac{9}{10})}^{2}} = \sqrt{169.9 - 0.81} = \sqrt{169.09} = 13$

$Now mean is \overset{―}{x} = A + \frac{Σ f_{i} d_{i}}{N} = 45 - (\frac{9}{10}) = 44.1$

$For Hashina$
$\begin{array}{lll} M a r k s (x_{i}) & d_{i} = x_{i} - 53 & d_{i}^{2} \\ 10 & - 43 & 1849 \\ 70 & 17 & 289 \\ 50 & - 3 & 9 \\ 20 & - 33 & 1089 \\ 95 & 42 & 1764 \\ 55 & 2 & 4 \\ 42 & - 11 & 121 \\ 60 & 7 & 49 \\ 48 & - 5 & 25 \\ 80 & 27 & 729 \\ TOTAL = 530 & = 0 & = 5928 \end{array}$
Here as $\frac{530}{10} = 53$ So, 53 is mean

And we know that the standard deviation can be written as,
$σ = \sqrt{\frac{Σ {(x - a)}^{2}}{n} - {(\frac{Σ (x - a)}{n})}^{2}}$

$\Rightarrow σ = \sqrt{\frac{5928}{10} - {(- \frac{0}{10})}^{2}} = \sqrt{592.8} = 24.35$

$We have the mean and S.D of the Hashina and Ravi.$

$For Ravi , C.V = \frac{σ}{\overset{―}{x}} \times 100 = \frac{13}{44.1} \times 100 = 29.48$
$For Hasina , C.V = \frac{σ}{\overset{―}{x}} \times 100 = \frac{24.35}{53} \times 100 = 45.94$

$Now as CV (of Ravi) < CV of Hasina$

$Hence, Ravi is more consistent$

$Mean of Hasina > Mean of Ravi$

$Hence, Hasina is more intelligent$

Question:22

Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Answer:

The mean and standard deviation of 100 observations were found to be 40 and 10, respectively.
If at the time of calculation, two observations were wrongly taken as 30 and 70 in place of 3 and 27
respectively.
We have to find the correct standard deviation
As per the given criteria, the Number of observations, n=100
Mean of the given observations before correction, $\overset{―}{X} = 40$

But we know $\overset{―}{X} = \frac{Σ x_{i}}{n}$

Substituting the corresponding values, we get $40 = \frac{Σ x_{i}}{100}$

$\Rightarrow Σ x_{i} = 40 \times 100 = 4000$

$It is given in the question that 2 observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively$

$So Σ x_{i} = 4000 - 30 - 70 + 3 + 27 = 3930$

$So the correct mean after correction is \overset{―}{X} = \frac{Σ x_{i}}{n} = \frac{3930}{100} = 39.3$

Also given that the standard deviation of the 100 observations is 10 before correction $σ = 10$

$But we know that σ = \sqrt{\frac{Σ x_{i}^{2}}{n} - {(\frac{Σ x_{i}}{n})}^{2}}$

$Substituting the corresponding values we get 10 = \sqrt{\frac{Σ x_{i}^{2}}{100} - {(\frac{4000}{100})}^{2}}$

$Now taking square on both sides, we get 10^{2} = \frac{Σ x_{i}^{2}}{100} - {(40)}^{2} 100 = \frac{Σ x_{i}^{2}}{100} - 1600 100 + 1600 = \frac{Σ x_{i}^{2}}{100} Σ x_{i}^{2} = 170000$

$It is said that 2 observations were wrongly taken as 30 and 70 in place of 3 and 27, respectively.$

$So, correction is Σ x_{i}^{2} = 170000 - {(30)}^{2} - {(70)}^{2} + {(3)}^{3} + {(27)}^{2}$

$Σ x_{i}^{2} = 170000 - 900 - 4900 + 9 + 729 = 164938$

$So, the correct standard deviation after correction is σ = \sqrt{\frac{164938}{100} - {(\frac{3930}{100})}^{2}} = \sqrt{1649.38 - {(39.3)}^{2}} = \sqrt{104.89} = 10.24$

Question:23

While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Answer:

$If at the time of calculation of mean and variance of 10 readings, a student wrongly used the reading 52 instead of the correct reading 25. The mean and variance were obtained as 45 and 16, respectively.$

$We have to find the correct mean and the variance.$

$As per the given criteria, Number of observations, n = 10 Mean of the given observations before correction, \overset{―}{X} = 45$

$But we know, \overset{―}{X} = \frac{Σ x_{i}}{n}$
$Substituting the corresponding values, we get 45 = \frac{Σ x_{i}}{10} Σ x_{i} = 45 * 10 = 450$

$It is given in the question that one reading 25 was taken wrongly as 52 S o Σ x_{i} = 450 - 52 + 25 = 423$

$So the correct mean after correction is \overset{―}{X} = \frac{Σ x_{i}}{n} = \frac{423}{10} = 42.3$
$Also given that variance of the 10 observations is 16 before correction σ^{2} = 16$

$But~we know that σ^{2} = \frac{Σ x_{i}^{2}}{n} - {(\frac{Σ x_{i}}{n})}^{2}$

$Substituting the corresponding values we get 16 = \frac{Σ x_{i}^{2}}{10} - {(45)}^{2}$

$Now taking square on both sides, we get 16 = \frac{Σ x_{i}^{2}}{10} - 2025 16 + 2025 = \frac{Σ x_{i}^{2}}{10} Σ x_{i}^{2} = 20410$
$It is said that one reading 25 was wrongly taken as 52. So,~correction is Σ x_{i}^{2} = 20410 - {(52)}^{2} + {(25)}^{2} = 20410 - 2704 + 625 = 18331$

$So, the variance after correction is σ^{2} = \frac{18331}{10} - {(\frac{423}{10})}^{2} = 1833.1 - {(42.3)}^{2} = 1833.1 - 1789.29 = 43.81$

Question:24

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
A. 2
B. 2.57
C. 3
D. 3.75

Answer:

Given data is 3, 10, 10, 4, 7, 10, 5. There are a total of 7.

$Here mean, \overset{―}{X} = \frac{3 + 10 + 10 + 4 + 7 + 10 + 5}{7} = \frac{49}{7} = 7$

$This can be written in table form as,$

$\begin{array}{ll} Data (x_{i}) & d_{i} = | x_{i} - \bar{x} | \\ 3 & 4 \\ 10 & 3 \\ 10 & 3 \\ 4 & 3 \\ 7 & 0 \\ 10 & 3 \\ 5 & 2 \\ Total & = 18 \end{array}$

$Hence Mean Deviation becomes, M . D = \frac{Σ d_{i}}{7} = \frac{18}{7} = 2.57$

Question:25

Mean deviation for n observations x₁, x₂, ..., x_n from their mean $\bar{X}$ is given by

$\begin{aligned} A. \sum_{i = 1}^{n} (x_{1} - \bar{x}) \\ B. \frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \overset{―}{x} | \\ C. \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} \\ D. \frac{1}{n} \sum_{i = 1}^{n} (x - \bar{x})^{2} \end{aligned}$

Answer:

$We know for n observations x_{1}, x_{2}, \dots \dots ., x_{n} h a v i n g \overset{―}{X} is given by M . D = \frac{Σ d_{i}}{n}$

$But we know that d_{i} = | x_{i} - \overset{―}{x} |$

$So, mean deviation becomes M.D = \frac{Σ | x_{i} - \overset{―}{x} |}{n}$

$\Rightarrow M.D = \frac{1}{n} \sum_{i = 0}^{n} | x_{i} - \overset{―}{x} |$

Question:26

When tested, the lives (in hours) of 5 bulbs were noted as follows:
1357, 1090, 1666, 1494, 1623
The mean deviations (in hours) from their mean is
A. 178
B. 179
C. 220
D. 356

Answer:

$Given that the lives of 5 bulbs in hours is 1357, 1090, 1666, 1494, 1623$

$Here mean \overset{―}{X} = \frac{1357 + 1090 + 1666 + 1494 + 1623}{5} = \frac{7230}{7} = 1446 This can be written in table form as,$
$\begin{array}{ll} Lives (in hours) x_{i} & d_{i} = x_{i} - \bar{x} \\ 1357 & 89 \\ 1090 & 356 \\ 1666 & 220 \\ 1494 & 48 \\ 1623 & 177 \\ Total & = 890 \end{array}$
Hence, mean deviation becomes

$M.D = \frac{Σ d_{i}}{n} = \frac{890}{5} = 178$

Question:27

Following are the marks obtained by 9 students in a mathematics test:
50, 69, 20, 33, 53, 39, 40, 65, 59
The mean deviation from the median is:
A. 9
B. 10.5
C. 12.67
D. 14.76

Answer:

Given that the marks obtained by 9 students in a mathematics test are 50, 69 , 20 ,33, 53, 39 ,40, 65, 59 As number of students =9 which is odd
$So median will be \frac{9 + 1}{2} = 5^{t h} t e r m$

$Arranging these in ascending order, we get 20,33,39,40,50,53,59,65,69$

$So the 5^{t h} term after arranging is 50 So, median is 50$

$This can be written in table form as,$
$\begin{array}{ll} Marks (x_{i}) & d_{i} =∣ x_{i} - Median ∣ \\ 20 & 30 \\ 33 & 17 \\ 39 & 11 \\ 40 & 10 \\ 50 & 0 \\ 53 & 3 \\ 59 & 9 \\ 65 & 15 \\ 69 & 19 \\ Total & = 114 \end{array}$
Hence Mean deviation becomes

$M . D = \frac{Σ d_{i}}{n} = \frac{114}{9} = 12.67$

Question:28

The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is

A. $\sqrt{\frac{52}{7}}$
B. $\frac{52}{7}$
C. $\sqrt{6}$
D. $6$

Answer:

Given data can be written in table form as,
$\begin{array}{ll} Data (x_{i}) & x_{i}^{2} \\ 6 & 36 \\ 5 & 25 \\ 9 & 81 \\ 13 & 169 \\ 12 & 144 \\ 8 & 64 \\ 10 & 100 \\ Total = 63 & 619 \end{array}$
$But we know that standard deviation can be written as σ = \sqrt{\frac{Σ x_{i}^{2}}{n} - {(\frac{Σ x_{i}}{n})}^{2}} = \sqrt{\frac{619}{7} - {(\frac{63}{7})}^{2}}$

$σ = \sqrt{\frac{619}{7} - {(9)}^{2}} = \sqrt{\frac{619}{7} - 81} = \sqrt{\frac{52}{7}}$

Question:29

Let x₁, x₂, ..., x_n be n observations and $\bar{X}$ be their arithmetic mean. The formula for the standard deviation is given by

$A . {(x_{i} - \overset{―}{x})}^{2} B . \frac{{(x_{i} - \bar{x})}^{2}}{n} C . \sqrt{\frac{{(X_{i} - \overset{―}{X})}^{2}}{n}} D \sqrt{\frac{x_{i}^{2}}{n} + {\bar{x}}^{2}}$

Answer:

$We know standard deviation for x_{1}, x_{2}, \dots . ., x_{n}$

$observations can be written as σ = \sqrt{\frac{1}{n} \sum_{i = 0}^{n} {(x_{i} - \overset{―}{x})}^{2}}$

Question:30

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is
A. 50000
B. 250000
C. 252500
D. 255000

Answer:

As per the question

Number of observations, n =100

$Mean of the given observations, \overset{―}{x} = 50$

$But we know, \overset{―}{X} = \frac{Σ x_{i}}{n}$

$Substituting the corresponding values, we get 50 = \frac{Σ x_{i}}{100} Σ x_{i} = 50 * 100 = 5000$

$It is also given that the standard deviation of the 100 observations is 5 σ = 5 σ = \sqrt{\frac{Σ x_{i}^{2}}{n} - {(\frac{Σ x_{i}}{n})}^{2}}$

$Substituting the corresponding values, we get 5 = \sqrt{\frac{Σ x_{i}^{2}}{100} - {(\frac{5000}{100})}^{2}}$

$Now taking square on both sides we get 5^{2} = \frac{Σ x_{i}}{100} - {(50)}^{2} 25 = \frac{Σ x_{i}}{100} - 2500 25 + 2500 = \frac{Σ x_{i}^{2}}{100} Σ x_{i}^{2} = 2525 * 100 Σ x_{i}^{2} = 252500$

Question:31

Let a, b, c, d, e be the observations with mean m and standard deviation s.

The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is
A. s
B. ks
C. s + k
D. $\frac{s}{k}$

Answer:

Given observations are a, b, c, d, e

So, the mean of the 5 observations is

$m = \frac{a + b + c + d + e}{5} = \frac{\sum x_{i}}{5}$

$\Rightarrow \sum x_{i} = 5 m = a + b + c + d + e$

And the standard deviation of the 5 observations is

$σ = \sqrt{\frac{\sum {(x_{i})}^{2}}{5} - {(\frac{\sum x_{i}}{5})}^{2}} = \sqrt{\frac{\sum {(x_{i})}^{2}}{5} - m^{2}}$

Now we will find the mean and standard deviation of the observations a+k , b+k, c+k, d+k, e+k we get

So, the mean of these 5 observations is

$m_{1} = \frac{(a + k) + (b + k) + (c + k) + (d + k) + (e + k)}{5} = \frac{a + b + c + d + e}{5} + k = m + k$

$σ_{1} = \sqrt{\frac{\sum {(x_{i} + k)}^{2}}{5} - {(\frac{\sum (x_{i} + k)}{5})}^{2}}$

$= \sqrt{\frac{\sum (x_{i}^{2} + k^{2} + 2 k x_{i})}{5} - {(m + k)}^{2}}$

$= \sqrt{\frac{\sum (x_{i}^{2})}{5} + \frac{5 k^{2}}{5} + \frac{2 k \sum (x_{i})}{5} - {(m + k)}^{2}}$

$= \sqrt{\frac{\sum (x_{i}^{2})}{5} + k^{2} + 2 m k - m^{2} - k^{2} - 2 m k}$

$= \sqrt{\frac{\sum (x_{i}^{2})}{5} - m^{2}}$

$\Rightarrow σ_{1} = σ$

Question:32

Let x₁, x₂, x₃, x₄, x₅ be the observations with mean m and standard deviation s.
The standard deviation of the observations kx₁, kx₂, kx₃, kx₄, kx₅ is
A. k + s
B. $\frac{s}{k}$
C. ks
D. s

Answer:

$Given observations are x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$

$So the mean of the 5 observations is$

$m = \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{5} = \frac{\sum x_{i}}{5}$

$\Rightarrow \sum x_{i} = 5 m = x_{1} + x_{2} + x_{3} + x_{4} + x_{5}$

$σ = \sqrt{\frac{\sum {(x_{i})}^{2}}{5} - {(\frac{\sum x_{i}}{5})}^{2}}$

$= \sqrt{\frac{\sum {(x_{i})}^{2}}{5} - m^{2}}$

$m_{1} = \frac{k x_{1} + k x_{2} + k x_{3} + k x_{4} + k x_{5}}{5}$

$= \frac{k (x_{1} + x_{2} + x_{3} + x_{4} + x_{5})}{5} = k m$
$σ_{1} = \sqrt{\frac{\sum {(k x_{i})}^{2}}{5} - {(\frac{\sum k x_{i}}{5})}^{2}} = \sqrt{\frac{\sum k^{2} {(x_{i})}^{2}}{5} - {(m_{1})}^{2}}$

$= \sqrt{\frac{k^{2} \sum {(x_{i})}^{2}}{5} - k^{2} m^{2}} = \sqrt{k^{2} (\frac{\sum {(x_{i})}^{2}}{5} - m^{2})} = k \sqrt{\frac{\sum {(x_{i})}^{2}}{5} - m^{2}} σ_{1} = k σ Hence, the standard deviation of new set of observations k x_{1}, k x_{2}, k x_{3}, k x_{4}, k x_{5} i s k σ$

Question:33

Let $x_{1}, x_{2}, . . . x_{n}$ be n observations. $w_{i} = l x_{i} + k$ for $i = 1, 2, . . . n,$ where l and k are constants. If the mean of $x_{i}$ 's is 48 and their standard deviation is 12, the mean of $w_{i}$ ’s is 55 and standard deviation of $w_{i}$ ’s is 15, the values of l and k should be
A. l = 1.25, k = – 5
B. l = – 1.25, k = 5
C. l = 2.5, k = – 5
D. l = 2.5, k = 5

Answer:

Given $x_{1}, x_{2} \dots . . x_{n} be n observations$

$And mean of these n observations, \overset{―}{x} = 48$

$And their standard deviation S D_{x} = 12$

$Another series of n observations is given such that w_{i} = l x_{i} + k f o r i = 1, 2, \dots . n where L and k are constants$

$And mean of these n observations, \overset{―}{w} = 55 And their standard deviation S D_{w} = 15$

$Applying the given condition for mean we get w_{i} = l x_{i} + k$

$Substituting the corresponding given values of means, we get 55 = l (48) + k \dots \dots (i)$
$We know that if standard deviation of x series is s, then standard deviation of kx series is ks$

$So standard deviation of x_{1}, x_{2}, \dots . x_{n} is S D_{x} ~ And hence the SD of l x_{1}, l x_{2} \dots . . l x_{n} is lS D_{x} Similarly, If standard deviation of x series is s, then standard deviation of k+x series is s,$

$So S.D of l x_{1}, l x_{2} \dots . l x_{n} is lS D_{x}$

$And hence the SD of l x_{1} + k, l x_{2} + k \dots . . l x_{n} + k i s l S D_{x}$

$So applying the given condition for standard deviation we get S D_{w} = l S D_{x}$

$Substituting the given values we get 15 = l (12) l = \frac{15}{12} = 1.25$

$Now substituting the value of l in equation (i), we get 55 = (1.25) (48) + k 55 = 60 + k k = - 5$

Question:34

Standard deviations for first 10 natural numbers is
A. 5.5
B. 3.87
C. 2.97
D. 2.87

Answer:

$We know that the standard deviation of the first n natural numbers is \sqrt{\frac{n^{2} - 1}{12}}$

$Now for first 10 natural numbers n = 10 substituting this in the equation of standard deviation we get σ = \sqrt{\frac{{(10)}^{2} - 1}{12}} = \sqrt{\frac{100 - 1}{12}} = \sqrt{\frac{99}{12}} = \sqrt{8.25} = 2.87$

Question:35

Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is
A. 6.5
B. 2.87
C. 3.87
D. 8.25

Answer:

We know that the standard deviation of the first n natural numbers is $\sqrt{\frac{n^{2} - 1}{12}}$
Now for first 10 natural numbers n=10
substituting this in the equation of standard deviation we get $σ = \sqrt{\frac{{(10)}^{2} - 1}{12}} = \sqrt{\frac{100 - 1}{12}} = \sqrt{\frac{99}{12}} = \sqrt{8.25} = 2.87$
Now when 1 is added to each numbers of 1 , 2, 3, 4, 5, 6,7 ,8 ,9, 10 ; we get new series as 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1
Now we know if standard deviation of x series is s, then standard deviation of k+x series is s,
So the S.D of 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1 series is also same as the S.D of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 series.
$σ = \sqrt{8.25}$
Now for variance we will square on both the sides $σ^{2} = 8.25$

Question:36

Consider the first 10 positive integers. If we multiply each number by –1 and then add 1 to each number, the variance of the numbers so obtained is
A. 8.25
B. 6.5
C. 3.87
D. 2.87

Answer:

First 10 positive integers are 1,2, 3, 4, 5,6 ,7, 8,9, 10 On multiplying each number by
-1 we get-1, -2, -3, -4 , -5, -6, -7, -8, -9 , -10
On adding 1 to each of the number, we get 0, -1, -2, -3, -4, -5, -6, -7, -8, -9
$Σ x_{i} = 0 - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 = - 45 a n d Σ x_{i}^{2} = 0^{2} + {(- 1)}^{2} + {(- 2)}^{2} + \dots \dots + {(- 9)}^{2}$

$But we know that Σ x_{i}^{2} = \frac{9 (9 + 1) (2 (9) + 1)}{6} = 285$

$σ = \sqrt{\frac{Σ x_{i}^{2}}{n} - {(\frac{Σ x_{i}}{n})}^{2}}$

$Substituting the corresponding values, we get σ = \sqrt{28.5 - {(- 4.5)}^{2}} = \sqrt{28.5 - 20.25} = \sqrt{8.25}$

$Now for variance both the sides are squared σ^{2} = 8.25$

Question:37

The following information relates to a sample of size 60:

$x^{2} = 18000,$ $x = 960.$
The variance is
A. 6.63
B. 16
C. 22
D. 44

Answer:

$We know that S.D can be written as σ = \sqrt{\frac{Σ x_{i}^{2}}{n} - {(\frac{Σ x_{i}}{n})}^{2}}$

$But given Σ x^{2} = 18000, Σ x = 960, N = 60,$

$substituting these corresponding values, we get σ = \sqrt{\frac{18000}{60} - {(\frac{960}{60})}^{2}}$

$= \sqrt{300 - {(16)}^{2}} = \sqrt{300 - 256} = \sqrt{44}$

$Now for variance we square both the sides σ^{2} = 44$

Question:38

Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is
A. 0
B. 1
C. 1.5
D. 2.5

Answer:

$Given coefficient of variation of two distributions are C V_{1} = 50 a n d C V_{2} = 60$

$And the arithmetic means are \overset{―}{x_{1}} = 30, \overset{―}{x_{2}} = 25$

$We know that coefficient of variation can be written as CV = \frac{σ}{\overset{―}{x}} \times 100$

$Now for first distribution, we have C V_{1} = \frac{σ_{1}}{\overset{―}{x_{1}}} \times 100$

$Substituting corresponding values, we get 50 = \frac{σ_{1}}{30} \times 100 σ_{1} = 15 \dots \dots . (i)$

$Now for second distribution, we have C V_{2} = \frac{σ_{2}}{\overset{―}{x_{2}}} \times 100$
$Substituting corresponding values, we get 60 = \frac{σ_{2}}{25} \times 100 σ_{2} = 15 \dots \dots . (i i)$

$So from both the equations we get that the difference of their standard deviation is 0$

Question:39

The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be
A. 81
B. 57
C. 36
D. 25

Answer:

$Given that the standard deviation of some temperature data in degree celcius, σ_{c} = 5$

$We know that C = \frac{5}{9} (F - 32) F = \frac{9 C}{5} + 32$

$Now we know that If S.D of x series is s, then S.D of kx series is ks.$

$So if SD of some temperature data is σ_{c} = 5$

$Then SD of some temperature data in \frac{9 C}{5}, \frac{9 C}{5} σ_{c} = 5 * \frac{9}{5} = 9$

$Similarly, If SD of x series is s, then SD of k+x series is s,$
$So SD of some temperature data in \frac{9 C}{5} σ_{c} = 9$

$And thus the SD of some temperature data in \frac{9 C}{5} + 32 w i l l b e σ_{F} = \frac{9 C}{5} σ_{C} = 9$

$Now for variance, we will square on both sides, we get σ_{F} = 9^{2} = 81$

Question:40

Fill in the blanks
Coefficient of variation = $\frac{. . .}{m e a n} \times 100$

Answer:

Standard Deviation
We know that the Coefficient of variation can be written as
$C . V = \frac{σ}{\overset{―}{x}} \times 100 w h e r e σ = standard deviation$

Question:41

Fill in the blanks
If $\bar{X}$ is the mean of n values of x, then $\sum_{i = 1}^{n} (x_{i} - \overset{―}{x})$ is always equal to _______.
If a has any value other than $\bar{X}$ , then $\sum_{i = 1}^{n} {(x_{i} - \overset{―}{x})}^{2}$ is _________ than $\sum {(X_{i} - a)}^{2}$

Answer:

Zero, less than Given x is mean of n values, then the sum of all n terms is denoted by $\sum_{i = 1}^{n} x_{i},$
so difference of both these is laways equal to zero i.e. $\sum_{i = 1}^{n} (x_{i} - \overset{―}{x}) = 0$
And square of the above equation is also equal to zero, so $\sum_{i = 1}^{n} {(x_{i} - \overset{―}{x})}^{2} = 0$
Now if "a" has the value other than x then $\sum_{i = 1}^{n} {(x_{i} - \overset{―}{x})}^{2} > 0$
$S o, \sum_{i = 1}^{n} {(x_{i} - \overset{―}{x})}^{2} < \sum_{i = 1}^{n} {(x_{i} - a)}^{2}$

$So, if a has any value other than \overset{―}{x}, t h e n \sum_{i = 1}^{n} {(x_{i} - \overset{―}{x})}^{2} ~is less than \sum_{i = 1}^{n} {(x_{i} - a)}^{2}$

Question:42

Fill in the blanks:
If the variance of a data is 121, then the standard deviation of the data is _______.

Answer:

11
We know the square root of variance is standard deviation $σ^{2} = 121$
Taking square root on both sides, we get $σ = \sqrt{121} = 11$
So, if the variance of a data is 121, then the Standard deviation of the data is 11.

Question:43

Fill in the blanks
The standard deviation of a data is ___________ of any change in origin but is _____ on the change of scale.

Answer:

Independent, dependent
Change of origin means some value has been added or subtracted in the observation. And we know the S.D does not change if any value is added or subtracted from the observations, So SD is independent of change in origin.
However, Standard deviation is only affected by a change in scale, that is, when some value is multiplied or divided to observations. Hence, the SD of any data is independent of any change in origin but dependent on any change of scale.

Question:44

The sum of the squares of the deviations of the values of the variable is _______ when taken about their arithmetic mean.

Answer:

Minimum
The sum of the squares of the deviations of the values of the variable is minimum or least when taken about their arithmetic mean.

Question:45

Fill in the blanks
The mean deviation of the data is _______ when measured from the median.

Answer:

Least
Mean deviation is sum of all deviations of a set of a data about the data’s mean. In addition, it is widely believed that the median is usually between the mean and the mode. So, the mean deviation of the data is least when measured from the median.

Question:46

Fill in the blanks
The standard deviation is _______ to the mean deviation taken from the arithmetic mean.

Answer:

Greater than or equal to
SD is the difference between the square of deviation of data about the mean and the square of the mean. In addition, the mean deviation is the sum of all deviations of a set of data from the data’s mean. Hence, the SD is greater than or equal to the mean deviation taken from the arithmetic mean.

Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

· Central Tendency

· Mean

· Median

· Mode

· Measure of Dispersion

· Range

· Quartile deviation

· Mean deviation

· Standard deviation

· Mean deviation for grouped data

· Mean deviation for ungrouped data

· Variance

· Analysis of Frequency Deviation

NCERT Solutions for Class 11 Maths: Chapter Wise

NCERT Solutions for Class 11 Mathematics Chapter 1: Sets

NCERT Solutions for Class 11 Mathematics Chapter 2: Relations and Functions

NCERT Solutions for Class 11 Mathematics Chapter 3: Trigonometric Functions

NCERT Solutions for Class 11 Mathematics Chapter 4: Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Mathematics Chapter 5: Linear Inequalities

NCERT Solutions for Class 11 Mathematics Chapter 6: Permutations and Combinations

NCERT Solutions for Class 11 Mathematics Chapter 7: Binomial Theorem

NCERT Solutions for Class 11 Mathematics Chapter 8: Sequences and Series

NCERT Solutions for Class 11 Mathematics Chapter 9: Straight Lines

NCERT Solutions for Class 11 Mathematics Chapter 10: Conic Sections

NCERT Solutions for Class 11 Mathematics Chapter 11: Introduction to Three-Dimensional Geometry

NCERT Solutions for Class 11 Mathematics Chapter 12: Limits and Derivatives

NCERT Solutions for Class 11 Mathematics Chapter 13: Statistics

NCERT Solutions for Class 11 Mathematics Chapter 14: Probability

Importance of NCERT Exemplar Class 11 Maths Solutions Chapter 15

By referring to the NCERT Exemplar Class 11 Maths chapter 15 solutions, students can easily understand the various concepts of Statistics. Statistics carries a significant number of marks according to the CBSE paper pattern, and the students should concentrate on important concepts like standard deviation and mean deviation, as they are frequently asked in the examination.
The step-wise and straightforward format of the NCERT exemplar solutions for Class 11 Maths chapter 15 makes it easy for the students to understand the most complex concepts of Statistics in a quick and easygoing manner. The pdf is exclusively created by our experts for students who find the concepts of Statistics difficult.
Students can easily score marks in this chapter by thoroughly studying a few topics that are commonly asked in the final examinations. The students will understand the concept in the simplest way possible by going through the NCERT Exemplar Class 11 Maths solutions chapter 15 Statistics.

NCERT Exemplar Class 11 Mathematics Chapters

NCERT Exemplar Class 11 Mathematics Chapter 1: Sets

NCERT Exemplar Class 11 Mathematics Chapter 2: Relations and Functions

NCERT Exemplar Class 11 Mathematics Chapter 3: Trigonometric Functions

NCERT Exemplar Class 11 Mathematics Chapter 4: Principles of Mathematical Induction

NCERT Exemplar Class 11 Mathematics Chapter 5: Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Mathematics Chapter 6: Linear Inequalities

NCERT Exemplar Class 11 Mathematics Chapter 7: Permutations and Combinations

NCERT Exemplar Class 11 Mathematics Chapter 8: Binomial Theorem

NCERT Exemplar Class 11 Mathematics Chapter 9: Sequence and Series

NCERT Exemplar for Class 11 Mathematics Chapter 10: Straight Lines

NCERT Exemplar Class 11 Mathematics Chapter 11: Conic Sections

NCERT Exemplar Class 11 Mathematics Chapter 12: Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Mathematics Chapter 13: Limits and Derivatives

NCERT Exemplar Class 11 Mathematics Chapter 14: Mathematical Reasoning

NCERT Exemplar Class 11 Mathematics Chapter 15: Statistics

NCERT Exemplar Class 11 Mathematics Chapter 16: Probability

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Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Exemplar Class 11 Maths Chapter 15?

NCERT Exemplar Class 11 Maths Chapter 15, "Statistics," focuses on measures of dispersion, including range, mean deviation, variance, and standard deviation, with a focus on both ungrouped and grouped data, and frequency distributions.

2. Why is Statistics important in Class 11 Mathematics?

Statistics is important in Class 11 Mathematics because it provides tools for analyzing and interpreting data, which is crucial for understanding various real-world phenomena and making informed decisions. It helps in understanding data distribution, measures of central tendency (mean, median, mode), and dispersion (variance, standard deviation).

3. What is the difference between variance and standard deviation in Statistics?

In statistics, variance measures the spread of data points around the mean, while standard deviation is the square root of the variance and provides a measure of spread in the same units as the original data.

4. How to prepare for Class 11 Maths Statistics chapter effectively?

To effectively prepare for Class 11 Maths Statistics, focus on understanding the concepts, practice various problems, and revise regularly, using NCERT solutions and other resources to reinforce your knowledge.

Also Read

NCERT Solutions for Exercise 15.2 Class 11 Maths Chapter 15 - Statistics

NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics

NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 2 Motion In a Straight Line

NCERT Solutions for Class 11 Biology Chapter 14 Breathing and Exchange of Gases

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

NCERT Class 11 Biology Chapter 13 Notes Photosynthesis In Higher Plants - Download PDF Notes

NCERT Class 11 Biology Chapter 10 Notes Cell Cycle And Cell Division- Download PDF

Waves Class 11th Notes - Free NCERT Class 11 Physics Chapter 15 Notes - Download PDF

Statistics Class 10th Notes - Free NCERT Class 10 Maths Chapter 14 Notes - Download PDF

Oscillations Class 11th Notes - Free NCERT Class 11 Physics Chapter 14 Notes - Download PDF

Probability Class 11th Notes - Free NCERT Class 11 Maths Chapter 16 notes - Download PDF

NCERT Exemplar Class 11 Biology Solutions Chapter 9 Biomolecules

NCERT Exemplar Class 11 Biology Solutions Chapter 4 Animal Kingdom

NCERT Exemplar Class 11 Biology Solutions Chapter 7 Structural Organisation in Animals

NCERT Exemplar Class 11 Biology Solutions Chapter 20 Locomotion and Movement

NCERT Exemplar Class 11 Biology Solutions Chapter 19 Excretory Products and their Elimination

NCERT Exemplar Class 11 Biology Solutions Chapter 22 Chemical Coordination and Integration

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

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