NCERT Solutions for Exercise 14.4 Class 9 Maths Chapter 14 - Statistics

# NCERT Solutions for Exercise 14.4 Class 9 Maths Chapter 14 - Statistics

Edited By Sumit Saini | Updated on Jul 06, 2022 02:44 PM IST

NCERT solutions for exercise 14.4 Class 9 Maths chapter 14- In this section, we are going to be discussing new topics and concepts. . As from the previous exercise i.e. exercise 14.3, we know that statistics consist of many important topics like frequency, range, and different types of graphs in a single chapter of mathematics. Studying and analyzing other things to this exercise 14.4 Class 9 Maths is an extension exercise of exercise 14.2 of the chapter which had numerical problems based on frequency and range. In this exercise 14.4, we will be studying concepts mean, median and mode, in this right from formula to how it is going to be applied we will be studying all its methods sequentially.

The NCERT book Class 9 Maths chapter 14 exercise 14.4 also covers numerical exercise apart from theory it takes new terms into account like questions based on observations, maximum and minimum predicted values which are like the importance to it. This NCERT syllabus Chapter contains only real-life examples which are important and will be discussed in NCERT Solutions for Class 9 Maths chapter 14 exercise 14.4. Along with Class 9 Maths chapter 1 exercise 14.4 the following exercises are also present.

## Statistics Class 9 Maths Chapter 14 Exercise: 14.4

2, 3, 4, 5, 0, 1, 3, 3, 4, 3

Find the mean, median and mode of these scores.

$Mean=\frac{Sum\ of\ observations}{Number \ of\ observations}$

Number of observations, n = 10

$\\\bar{X}=\frac{\sum_{i=1}^{n=10}x_{i}}{n}\\ \bar{X}=\frac{ 2+ 3+ 4+ 5+ 0+ 1+ 3+ 3+ 4+ 3}{10}\\ \bar{X}=\frac{28}{10}\\ \bar{X}=2.8$

Mean is 2.8

To find the median we have to arrange the given data in ascending order as follows:

0, 1, 2, 3, 3, 3, 3, 4, 4, 5

n = 10 (even)

$\\Median=\frac{(\frac{n}{2})^{th}\ term+(\frac{n}{2}+1)^{th}\ term}{2}\\ Median=\frac{(\frac{10}{2})^{th}\ term+(\frac{10}{2}+1)^{th}\ term}{2}\\ Median=\frac{(5)^{th}\ term+(6)^{th}\ term}{2}\\ Median=\frac{3+3}{2}\\ Median=3$

In the given data 3 occurs the maximum number of times (4)

Therefore, Mode = 3

41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60

Find the mean, median and mode of this data.

$Mean=\frac{Sum\ of\ observations}{Number \ of\ observations}$

Number of observations, n = 15

$\\\bar{X}=\frac{\sum_{i=1}^{n=15}x_{i}}{n}\\ \bar{X}=\frac{41+ 39+ 48+ 52+ 46+ 62+ 54+ 40+ 96+ 52+ 98+ 40+ 42+ 52+ 60}{15}\\ \bar{X}=\frac{822}{15}\\ \bar{X}=54.8$

Mean is 54.8

To find the median we have to arrange the given data in ascending order as follows:

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

n = 15 (odd)

$\\Median=(\frac{n+1}{2})^{th}\ term\\ Median=(\frac{15+1}{2})^{th}\ term\\ Median=8^{th}\ term\\ Median=52$

In the given data 52 occurs the maximum number of times ()

Therefore, Mode = 52

29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

The given data is already in ascending order

Number of observations, n = 10 (even)

$\\Median=\frac{(\frac{n}{2})^{th}\ term+(\frac{n}{2}+1)^{th}\ term}{2}\\ Median=\frac{(\frac{10}{2})^{th}\ term+(\frac{10}{2}+1)^{th}\ term}{2}\\ Median=\frac{(5)^{th}\ term+(6)^{th}\ term}{2}\\ Median=\frac{(x)+(x+2)}{2}\\ Median=\frac{2x+2}{2}\\ Median=x+1$

x + 1 = 63

x = 62

In the given data 14 is occuring the maximum number of times (4)

Mode of the given data is therefore 14.

 Salary (in Rs) Number of workers 3000 16 4000 12 5000 10 6000 8 7000 6 8000 4 9000 3 10000 1 Total 60

 Salary ( in Rs)(x i ) Number of workers(f i ) f i x i 3000 16 48000 4000 12 48000 5000 10 50000 6000 8 48000 7000 6 42000 8000 4 32000 9000 3 27000 10000 1 10000 Total $\sum_{i=1}^{8}f_{i}=60$ $\sum_{i=1}^{8}f_{i}x_{i}=305000$

The mean of the above data is given by

$\\\bar{X}=\frac{\sum_{i=1}^{8}f_{i}x_{i}}{\sum_{i=1}^{8}f_{i}}\\ \bar{X}=\frac{305000}{60}\\ \bar{X}=5083.33$

The mean salary of the workers working in the factory is Rs 5083.33

The mean is an appropriate measure of central tendency in case the observations are close to each other. An example of such a case is height of the students in a class.

Statistics Excercise: 14.4

The mean is not an appropriate measure of central tendency in case the observations are not close to each other. An example of such a case is prices of the toys in a toy shop.

## More About NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4

Exercise 14.4 Class 9 Maths including problems based on smaller one and bigger ones, as per student’s methodology. The questions count increases and students gets an understanding of all the terminologies discussed above serial wise. Questions have been serialized according to the steps shown in NCERT solutions for Class 9 Maths chapter 14 exercise 14.4. In the lasts of Class 9 Maths chapter 14 exercise, 14.4 gives the important questions over all the topics, parts and examples of the chapter. The Class 9 Maths chapter 14 exercise 14.4 gives attributes to mean median and mode.

Also Read| Statistics Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4

• The very first benefit in completing exercise 14.4 Class 9 Maths is right from all exercises covered in the chapter, it is also a favourite topic of an examiner to frame question form

• Exercise 14.4 contains Questions related to the frequency and distributed data attached to Mean and median and questions related to grouping and ungrouping of data.

• Students can have a better understanding in exercise 14.4 Class 9 Maths getting concepts over it will give an edge over others with best out of it as mentioned Class 9 Maths chapter 14 exercise 14.4

• Class 9 Maths Chapter 14 exercise 14.4 is just a trailer for best problems to cover in future classes

Also, see-

## NCERT Solutions of Class 10 Subject Wise

1. Define Arithmetic Mean in statistics?

Ans: In mathematics and especially in statistics, the idea of mean is crucial. In a group of numbers, the mean is the average or most frequent value.

2. What do you mean by Median?

Ans: The median of a dataset listed in ascending order is a statistical metric that determines the middle value (i.e., from smallest to largest value)...

3. Can you explain Mode?

Ans:  In data collection, the model is the value that appears the most frequently. A collection of data might have only one mode, several modes, or none at all.

4. What is the formula of Mean for discrete data?

Ans:   .mean = fixi/xi where fi = frequency of discrete data and xi = discrete data

5. What is the difference between mean and median?

Ans: When a set of data is sorted in ascending or descending order, the mean is the average value, and the median is the midway value.

6. The number that appears the most frequently in a group of numbers is?

Ans: The number that appears the most frequently in a group of numbers is called MODE

7. The mode of 12, 17, 16, 15, 13, 11, 15 is?

Ans: The number 15 occurs the most(2 times) hence mode is 15 according to the definition.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

Exam Date:19 September,2024 - 19 September,2024

Exam Date:20 September,2024 - 20 September,2024

Exam Date:26 September,2024 - 26 September,2024

Application Date:30 September,2024 - 30 September,2024

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 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.

With increase of temperature, which of these changes?

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

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 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