NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability

Question:1

In the formula $\overline{x}=a+\frac{\sum f_i{d}_i}{\sum f_i}$
For finding the mean of grouped data, $d_i$’s are deviations from
(A) lower limits of the classes
(B) upper limits of the classes
(C) midpoints of the classes
(D) frequencies of the class marks

Answer: [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide
by the total number of observations.
We know that d_i = x_i – a
where x_i is data and a is the mean
So, $d_i$ is the deviation from the midpoint of the classes.

Question:2

While computing the mean of grouped data, we assume that the frequencies are
(A) evenly distributed over all the classes
(B) centred at the class marks of the classes
(C) centred at the upper limits of the classes
(D) centred at the lower limits of the classes

Answer: [B]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide by the total number of observations.
Hence, while computing the mean of grouped data, we assume that the frequencies are centered at the class marks of the classes

Question:3

If x_i’s are the mid points of the class intervals of grouped data, fi’s are the corresponding frequencies and $\overline{x}$ is the mean, then $\mathrm{\Sigma }(f_i{x}_i-\overline{x})$ is equal to
(A) 0 (B) –1 (C) 1 (D) 2

Answer: [A]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide by the total number of observations.
That is mean $\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{n}$
By cross multiplication, we get
$\sum f_i{x}_i=\overline{x}n\cdots (1)$
$\sum (f_i{x}_i-\overline{x})=\sum f_i{x}_i-\sum \overline{x}$
$=n\overline{x}-\sum \overline{x}$ (from equation (1))
$=n\overline{x}-n\overline{x}$
= 0

Question:4

In the formula $\overline{x}=a+h\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)$ for finding the mean of grouped frequency distribution, u_i =
(A) $\frac{x_i+a}{h}$ (B) h(xi – a) (C) $\frac{x_i-a}{h}$ (D) $\frac{a-x_i}{h}$

Answer:

Answer. [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide by the total number of observations.
Also we know that d_i = x_i – a and $u_i=\frac{d_i}{h}$
$u_i=\frac{d_i}{h}$
put d_i = x_i – a
$u_i=\frac{x_i-a}{h}$
Hence, option C is correct.

Question:5

The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(A) mean
(B) median
(C) mode
(D) all three above

Answer: [B]
Solution: Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
If we make a graph of less than type and of more than type grouped data and find the intersection point, then the value at the abscissa is the median of the grouped data.
Hence, option (B) is correct.

Question:6

For the following distribution:

The sum of the lower limits of the median class and the modal class is
(A) 15
(B) 25
(C) 30
(D) 35

Answer: [B]
Solution: Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

$\frac{N}{2}=\frac{66}{2}=33$ which ies in the class 10-15.
Hence, the median class is 10 – 15
The class with the maximum frequency is the modal class, which is 15 – 20
The lower limit of the median class = 10
The lower limit of the modal class = 15
Sum = 10 + 15 = 25

Question:7

Consider the following frequency distribution:

The upper limit of the median class is
(A) 17 ) 17.5 (C) 18 (D) 18.5

Answer: [B]

Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Class is not continuous. So we have to make 1t continuous first.

$\frac{N}{2}=\frac{57}{2}=28\cdot 5$
Here, the median class is 15.5 – 17.5
The upper limit of the median class is 17.5.

Question:8

For the following distribution :

The modal class is
(A) 10-20 (B) 20-30 (C) 30-40 (D) 50-60

Answer:

Answer. [C]
Solution.
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However, mostly we use frequency distribution to summarize categorical variables.

The class with the highest frequency is 30-40
Hence, 30 – 40 is the modal class.

Question:9

Consider the data:

The difference of the upper limit of the median class and the lower limit of the modal class is
(A) 0 (B) 19 (C) 20 (D) 38

Answer: [C]

Solution: Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

$\frac{N}{2}=\frac{67}{2}=33\cdot 5$
Median class = 125 – 145
upper limit of median = 145
The class with the maximum frequency is the modal class, which is 125 – 145
lower limit of modal class = 125
Difference of the upper limit of median and lower limit of modal = 145 – 125 = 20

Question:10

The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below :

The number of athletes who completed the race in less than 14.6 seconds is :
(A) 11 (B) 71 (C) 82 (D) 130

Answer: [C]
Solution: Frequency:- The number of times an event occurs in a specific period is called frequency.
The number of athletes who are below 14.6 = frequency of class (13.8-14) + frequency of class (14- 14.2) +
frequency of class (14.2-14.4) + frequency of class (14.4-14.6)
= 2 + 4 + 5 + 71 = 82
Hence, the frequency of race completed in less than 14.6 = 82

Question:11

Consider the following distribution :

The frequency of the class 30-40 is
(A) 3 (B) 4 (C) 48 (D) 51

Answer: [A]
Solution: Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

So, the frequency of class 30 – 40 is 3.

Question:12

If an event cannot occur, then its probability is
(A) 1 (B)$\frac{3}{4}$ (C) $\frac{1}{2}$ (D) 0

Answer. [D]
Solution: Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Here number of favorable cases is 0.
Probability = $\frac{Numberoffavourablecase}{Totalnumberofcases}$
$\Rightarrow$Probability = $\frac{0}{\left(Totalcases\right)}=0$

Question:13

Which of the following cannot be the probability of an event?
(A)$\frac{1}{3}$ (B) 0.1 (C) 3% (D)$\frac{17}{16}$

Answer:

Answer. [D]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
(A) 1/3
Here 0 < 1/3 < 1
Hence, it can be the probability of an event.
(B) 0.1
Here 0 < 0.1 < 1
Hence, it can be the probability of an event.
(C) 3% = 3/100 = 0.03
Here 0 < 0.03 < 1
Hence, it can be the probability of an event.
(D)17/16
Here $\frac{17}{16}>1$
Hence $\frac{17}{16}$ is not a probability of event
Hence, option (D) is the correct answer.

Question:14

An event is very unlikely to happen. Its probability is closest to
(A) 0.0001 (B) 0.001 (C) 0.01 (D) 0.1

Answer:

Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The descending order of option (A), (B), (C), (D) is
0.1 > 0.01 > 0.001 > 0.0001 that is (D) > (C) > (B) > (A)
We can also say that it is the order of happening of an event.
Here, 0.0001 is the smallest one.
Hence, 0.0001 is very unlikely to happen

Question:15

If the probability of an event is p, the probability of its complementary event will be
(A) p – 1 (B) p (C) 1 – p (D)$1-\frac{1}{p}$

Answer:

Answer. [C]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of an event = p
Let the probability of its complementary event = q
We know that total probability is equal to 1.
Hence, p + q = 1
$\Rightarrow$q = 1 – p

Question:16

The probability expressed as a percentage of a particular occurrence can never be
(A) less than 100
(B) less than 0
(C) greater than 1
(D) anything but a whole number

Answer:

Answer. [B]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability expressed as a percentage of an event A is between 0 to 100.
Hence, we can say that probability can never be less than 0.
Hence, option (B) is correct.

Question:17

If P(A) denotes the probability of an event A, then
(A) P(A) < 0 (B) P(A) > 1 (C) 0 ≤ P(A) ≤ 1 (D) –1 ≤ P(A) ≤ 1

Answer:

Answer. [C]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
(A) P(A) < 0
It does not represent the probability of event A because the probability of an event can never be less than 0.
(B) P(A) > 1
It does not represent the probability of event A because the probability of an event can never be greater than 1.
(C) 0 ≤ P(A) ≤ 1
It represents the probability of event A because the probability of an event always lies between 0 to 1.
(D) –1 ≤ P(A) ≤ 1
It does not represent the probability of event A because the probability of an event can never be equal to -1.
Hence, option (C) is correct.

Question:18

A card is selected from a deck of 52 cards. The probability of its being a red face card is
(A) $\frac{3}{26}$ (B) $\frac{3}{13}$ (C) $\frac{2}{13}$ (D) $\frac{1}{2}$

Answer:

Answer. [A]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 52
Red face cards = 6
Favorable cases = 6
Let event A is to selection of a card from 52 cards.
The probability that it is a red card is p(A)
$P\left(A\right)=\frac{Numberoffavourablecases}{Totalnumberofcases}$
$\Rightarrow$$P\left(A\right)=\frac{6}{52}=\frac{3}{26}$

Question:19

The probability that a non-leap year selected at random will contain 53 Sundays is
(A) $\frac{1}{7}$ (B) $\frac{2}{7}$ (C) $\frac{3}{7}$ (D) $\frac{5}{7}$

Answer:

Answer. [A]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
In 365 days, there are 52 weeks and 1 day.
If it contains 53 Sundays, then the 1 day of the year must be Sunday.
But there are a total of 7 days.
Hence, the total number of favourable cases = 1
Hence probability of 53 sunday = $\frac{Numberoffavourablecases}{Totalcases}$
$=\frac{1}{7}$

Question:20

When a die is thrown, the probability of getting an odd number less than 3 is
(A) $\frac{1}{6}$ (B) $\frac{1}{3}$ (C) $\frac{1}{2}$ (D) 0

Answer:

Answer. [A]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total no. of cases = 6
odd number less than 3 = 1
Number of favorable cases = 1
Probability = $=\frac{1}{6}$

Question:21

A card is drawn from a deck of 52 cards. The event E is that the card is not an ace of hearts. The number of outcomes favourable to E is
(A) 4 (B) 13 (C) 48 (D) 51

Answer:

Answer. [D]
Solution. Total number of cards = 52
Ace of hearts = 1
The card is not an ace of hearts = 52 – 1 = 51
The number of outcomes favourable to E = 51

Question:22

The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is
(A) 7 (B) 14 (C) 21 (D) 28

Answer:

Answer. [B]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Let event A be to get a bad egg.
So, p (A) = 0.035 (given)
P(A) = $\frac{Numberoffavorablecases}{Totalnumberofcases}$
$\Rightarrow$0.035 = $\frac{Numberoffavorablecases}{400}$
Number of favourable cases = $\frac{35}{1000}\times 400=\frac{140}{10}=14$

Question:23

A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought?
(A) 40 (B) 240 (C) 480 (D) 750

Answer:

Answer. [C]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases = 6000
Probability of getting first prize (p(A)) = 0.08
p(A) $=\frac{Numberofticketsthebought}{Totaltickets}$
$\Rightarrow$0.08 × 6000 = Number of tickets the bought
Number of tickets they bought = 480.

Question:24

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is
(A) $\frac{1}{5}$ (B) $\frac{3}{5}$ (C) $\frac{4}{5}$ (D) $\frac{1}{3}$

Answer:

Answer. [A]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total tickets = 40
Number of tickets multiple of 5 = 5, 10, 15, 20, 25, 30, 35, 40
Total favourable cases = 8
Let A be the event of getting a ticket with a number multiple of 5.
p(A) = $\frac{Numberofticketsthebought}{Totaltickets}$
$\Rightarrow$$p\left(A\right)=\frac{8}{40}=\frac{1}{5}$

Question:25

Someone is asked to take a number from 1 to 100. The probability that it is a prime is
(A) $\frac{1}{5}$ (B) $\frac{6}{25}$ (C) $\frac{1}{4}$ (D) $\frac{13}{50}$

Answer:

Answer. [C]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 100
prime number from 1 to 100 = 2, 3, 5, 7, 9, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 97
Total prime numbers from 1 to 100 = 25
Probability of getting prime number = $\frac{primeno.from1to100}{Totalnumber}$
$$\begin{gathered} =\frac{25}{100} \\ =\frac{1}{4} \end{gathered}$$

Question:26

A school has five houses: A, B, C, D, and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D, and the rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B, and C is
(A) $\frac{4}{23}$ (B) $\frac{6}{23}$ (C) $\frac{8}{23}$ (D) $\frac{17}{23}$

Answer:

Answer. [B]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total students = 23
Students in A, B, C = 4 + 8 + 5 = 17
Students in C, D = 23 – 17 = 6
Number of favourable cases = 6
Let A be the event that the student is not from A, B, C
$p\left(A\right)=\frac{StudentfromC,D}{Totalstudents}$
$\Rightarrow$$p\left(A\right)=\frac{6}{23}$

Question:1

The median of ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give a reason.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Ungrouped data:- The data that is not grouped is called ungrouped data.
The median is the middle number in the grouped data, but when the data is ungrouped, the median is also changed.
Hence, the median is not the same for grouped and ungrouped data

Question:2

In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula
$\overline{x}=a+\frac{\sum f_i{d}_i}{\sum f_i}$where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Mean: It is the average of the given numbers. It is easy
to calculate the mean. First of all, add up all the numbers, then divide by how many numbers there are.
This last statement is not correct because a can be any point in the grouped data; it is not necessary that a must be the midpoint.
Hence, the statement is false.

Question:3

Is it true to say that the mean, mode, and median of grouped data will always be different? Justify your answer.

Answer:

Answer. [False]
Solution. Mean: It is the average of the given numbers. It is easy to calculate the mean. First of all, add up all the numbers, then divide by how many numbers are there.
Grouped data are data formed by aggregating individual observations of a variable into groups.
The mean, mode, and median of grouped data can be the same it will depend on what type of data is given.
Hence, the statement is false.

Question:4

Will the median class and modal class of grouped data always be different? Justify your answer.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
The median is always the middle number, and the modal class is the class with the highest frequency it can happen that the median class is of the highest frequency.
So the given statement is false, the median class and mode class can be the same.

Question:5

In a family having three children, there may be no girls, one girl, two girls or three girls. So, the probability of each is $\frac{1}{4}$. Is this correct? Justify your answer.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total children = 3
Cases – GGG, GGB, GBG, BGG, BBB, BBG, BGB, GBB were G is girl and B is boy.
Probability = $\frac{Numberoffavorablecases,}{Totalnumberofcases}$
Probability of 0 girl = $\frac{1}{8}$
Probability of 1 girl = $\frac{3}{8}$
Probability of 2 girl = $\frac{3}{8}$
Probability of 3 girl = $\frac{1}{8}$
Here they are not equal to $\frac{1}{4}$

Question:6

A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig.). Are the outcomes 1, 2, and 3 equally likely to occur? Give reasons

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here, 3 contains 50% of the region, and 1, 2 contain 25%, 15% of the region.
Probability= $\frac{Numberoffavorablecases}{Totalnumberofcases}$
Probability of 3= $\frac{50}{100}=\frac{1}{2}$
Probability of 1= $\frac{25}{100}=\frac{1}{4}$
probability of 2= $\frac{25}{100}=\frac{1}{4}$
All probabilities are not equal. So the given statement is false.

Question:7

Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why

Answer:

Answer. [Peehu]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
As Apoorv throws two dice total cases = 36
Product is 36 when he get = (6, 6)
Number of favourable cases = 1
Probability = $\frac{Numberoffavourablecases}{Totalnumberofcases}$
Probability that Apoorv get 36 = $\frac{1}{36}$
Peehu throws are die total cases = 6
The square of 6 is 36
Hence case = 1
Probability that Peehu get 36 = $\frac{1}{6}$
Hence, Peehu has better cases to get 36.

Question:8

When we toss a coin, there are two possible outcomes - Head or Tail. Therefore, the probability of each outcome is $\frac{1}{2}$. Justify your answer.

Answer:

Answer. [True]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases when we toss a coin = 2(H, T)
Probability = $\frac{Numberoffavourablecases,}{Totalnumberofcases}$
Probability of head = $\frac{1}{2}$
Probability of tail = $\frac{1}{2}$
Hence, the probability of each outcome is $\frac{1}{2}$.

Question:9

A student says that if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and
the probability of getting ‘not 1’ each is equal to $\frac{1}{2}$. Is this correct? Give reasons.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here total cases = 6
Number of favourable cases in getting 1 = 1
Probability = $\frac{Numberoffavourablecases,}{Totalnumberofcases}$
Probability of getting $1=\frac{1}{6}$
Number of favourable cases 'not 1' = 5 (2, 3, 4, 5, 6)
Probability of not 1 = $\frac{5}{6}$
Hence, they are not equal to $\frac{1}{2}$

Question:10

I toss three coins together. The possible outcomes are no heads, 1 head, 2 heads, and 3 heads. So, I say that the probability of no heads is $\frac{1}{4}$. What is wrong with this conclusion?

Answer:

Answer. [$\frac{1}{8}$ ]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases in tossing three coins = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Number of cases with no head = TTT
Probability = $\frac{Numberoffavourablecases}{Totalnumberofcases}$
Probability of no head =$\frac{1}{8}$
The conclusion that the probability of no head is $\frac{1}{4}$ is wrong because, as we calculated above, it comes out to $\frac{1}{8}$. Hence the probability of no head is $\frac{1}{8}$

Question:11

If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1? Give reasons.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of getting a head is 1, which means that we never get a tail. But this is not true because we have both heads and tails on a coin. Hence probability of getting a head is 1 is false.

Question:12

Sushma tosses a coin 3 times and gets tails each time. Do you think that the outcome of the next toss will be a tail? Give reasons.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because when we toss a coin, we can get either tail or head, and the probability of each is $\frac{1}{2}$.
So, it is not necessary that she gets tails on the fourth toss. She can also get a head.

Question:13

If I toss a coin 3 times and get heads each time, should I expect a tail to have a higher chance in the 4th toss? Give a reason in support of your answer.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because we get head or tail after tossing a coin, the probability of both outcomes is $\frac{1}{2}$ .
Hence tail does not have a higher chance than the head.
Both have an equal chance.

Question:14

A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, so, the probability of each is $\frac{1}{2}$. Justify.

Answer:

Answer. [True]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total slips = 100
Slips with an even number = 50
Probability = $\frac{Numberoffavourablecases}{Totalnumberofcases}$
Probability of slip with even number = $\frac{50}{100}=\frac{1}{2}$
Slips with odd number = 50
Probability of slip with odd number = $\frac{50}{100}=\frac{1}{2}$
Hence, the probability of each is $\frac{1}{2}$.

Question:1

Find the mean of the distribution:

Answer:

Answer. [5.5]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

   Class	Marks (xi)	Frequency(fi)	fixi
   1-3	2	9	18
   3-5	4	22	88
   5-7	6	27	162
   7-10	8.5	17	144.5
   		$\sum f_i=75$	$\sum f_i{x}_i=412\cdot 5$

   
   $mean\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{412\cdot 5}{75}=5\cdot 5$

Question:2

Calculate the mean of the scores of 20 students in a mathematics test :

Answer:

Answer. [35]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)
   4 Now divide the sum of (fx) by the sum of (f)

$\sum f_i=20$ $\sum f_i{x}_i=700$
$mean\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{700}{20}=35$

Question:3

Calculate the mean of the following data

Answer:

Answer. [12.93]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

$mean\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{362}{28}=12\cdot 93$

Question:4

The following table gives the number of pages written by Sarika for completing her own book for 30 days :

Find the mean number of pages written per day.

Answer:

Answer. [26]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and sthe um of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

   No.of pages written per day	no.of days(fi)	(xi)	fixi
   16-18	1	17	17
   19-21	3	20	60
   22-24	4	23	92
   25-27	9	26	234
   28-30	13	29	377
   	$\sum f_i=30$		$\sum f_i{x}_i=780$

   
   $mean\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{780}{30}=26$

Question:5

The daily income of a sample of 50 employees are tabulated as follows :

Find the mean daily income of employees.

Answer:

Answer. [356.5]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

   Income (in Rs )	xi	No.of employees	fixi
   1-200	100.5	14	1407
   201-400	300.5	15	4507.5
   401-600	500.5	14	7007
   601-800	700.5	7	4903.5
   		$\sum f_i=50$	$\sum f_i{x}_i=17825$

   $mean\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{17825}{50}=356.5$

Question:6

An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table :

Determine the mean number of seats occupied over the flights.

Answer:

Answer. [109]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

$mean\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{10992}{100}=109\cdot 92$
number of seats = 109

Question:7

The weights (in kg) of 50 wrestlers are recorded in the following table :

Find the mean weight of the wrestlers.

Answer:

Answer. [123.4]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

   Weight	fi	xi	fixi
   100-110	4	105	420
   110-120	14	115	1610
   120-130	21	125	2625
   130-140	8	135	1080
   140-150	3	145	435
   	$\sum f_i=50$		$\sum f_i{x}_i=6170$

   
   $mean\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{6170}{50}=123\cdot 4kg$

Question:8

The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below :

Find the mean mileage.
The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?

Answer:

Answer. [14.48]
Solution. Here we calculate the mean by following
1. Find the midpoint of each interval.
2. Multiply the frequency of each interval by its midpoint.
3. Get the sum of all the frequencies (f) and the sum of all the (fx)
4. Now divide the sum of (fx) by the sum of (f)

$mean\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{724}{50}=14\cdot 48km/L$

Question:9

The following is the distribution of weights (in kg) of 40 persons :

Construct a cumulative frequency distribution (of the less than type) table for the data above.

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However, mostly we use frequency distribution to summarize categorical variables.

Question:10

The following table shows the cumulative frequency distribution of marks of 800 students in an examination:

Construct a frequency distribution table for the data above.

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed over values in a
frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Question:11

Form the frequency distribution table from the following data :

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Question:12

Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students in a class :

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to
summarize categorical variables.
$a=12$ ( because first term of frequency and cumulative frequency is same )
$\Rightarrow$12 + b = 25
$\Rightarrow$b = 25 – 12
$\Rightarrow$b = 13
$\Rightarrow$25 + 10 = c
$\Rightarrow$35= c
$\Rightarrow$c + d = 43
$\Rightarrow$35 + d = 43
$\Rightarrow$d = 43 – 35
$\Rightarrow$d=8
$\Rightarrow$43 + e = 48
$\Rightarrow$e = 48 – 43
$\Rightarrow$e =5
$\Rightarrow$48+2 = f
$\Rightarrow$50 = f
Ans. a = 12, b = 13, c = 35, d = 8, e = 5, f = 50

Question:13

The following are the ages of 300 patients getting medical treatment in a hospital on a particular day :

(i) Less than type cumulative frequency distribution.
(ii) More than type cumulative frequency distribution

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Question:14

Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class :