 # T.R. Jain and V.K. Ohri Solutions Class 11 Statistics Economics Chapter 9 - Measures of Central Tendency- Arithmetic Mean

T.R. Jain and V.K. Ohri Solutions for class 11 statistics for economics chapter 9: Measures of central tendency: Arithmetic mean, is regarded as an important concept to be studied thoroughly by the students. Here, we have provided T.R. Jain and V.K. Ohri solutions for class 11.

 Board CBSE Class Class 11 Subject Statistics for Economics Chapter Chapter 9 Chapter Name Measures of Central Tendency- Arithmetic Mean Number of questions solved 05 Category T.R. Jain and V.K. Ohri

This chapter covers the following mentioned concepts:

• Good average
• Arithmetic mean
• Types of statistical averages
• Simple arithmetic mean

## T.R. Jain and V.K. Ohri Solutions for Class 11 Statistics for Economics Chapter 9 – Measures of Central Tendency- Arithmetic Mean

Question 1

Following are the marks obtained by eight students in statistics. Calculate the arithmetic mean.

 Marks 15 18 16 45 32 40 30 28

Solution.

 Marks (X) 15 18 16 45 32 40 30 28 X= 224

$$\begin{array}{l}{\bar{X}}\, =\, \frac{\sum X}{N}\, =\,\frac{X_{1}+X_{2}+…+X_{10}}{10}\, =\frac{224}{8}\, =28\end{array}$$

Average marks of the eight students = 28

Question 2

A train runs for 25 miles at a speed of 30 mph, another for 50 miles at a speed of 40 mph. Due to repairs of the track, the trains runs for 6 minutes at a speed of 10 mph and finally covers the remaining distance of 24 miles at a speed of 24 mph. What is the average speed in miles per hour?

Solution.

Time taken in covering 25 miles as at speed of 30 mph = 50 minutes

( ∵ Time =

$$\begin{array}{l}\frac{Distance}{Speed}\end{array}$$
)

Time taken in covering 50 miles at a speed of 40 mph = 75 minutes

Distance covered in 6 minutes at a speed of 10 mph = 1 mile

Time taken in covering 24 miles at speed of 24 mph = 60 minutes

Therefore, taking the time taken as weights we have the weighted mean as:

 Speed in mph (X) Time taken (W) WX 30 40 10 24 50 75 6 60 1,500 3,000 60 1,440 ∑W= 191 ∑WX= 6,000

Weighted Mean,

$$\begin{array}{l}{\bar{X}}_{w}\, =\frac{\sum WX}{\sum W}\, =\frac{6,000}{191}=31.41\end{array}$$

Average Speed= 31.41 mph.

Question 3

In a class of 50 students, 10 have failed and their average of marks is 2.5. The total marks secured by the entire class were 281. Find the average marks of those who have passed.

Solution.

Given N = 50, failed students = 10

Mean marks of those who failed = 2.5

Total marks of 10 students who have failed = 2.5 x 10 = 2.5

Total marks secured by the entire class = 281

Total marks obtained by those who have passed = 281 – 25 = 256

Average marks obtained by those who have passed =

$$\begin{array}{l}\frac{256}{40}\end{array}$$
= 6.4

Average marks obtained by those who have passed = 6.4

Question 4

Calculate the mean marks from the following data.

 Marks 20-25 25-30 30-35 35-40 40-45 45-50 50-55 Number of students 10 12 8 20 11 4 5

Solution

 Marks (X) Mid-value m= $$\begin{array}{l}\frac{l_{1}+l_{2}}{2}\end{array}$$ Number of students or frequency (f) Deviation (d=m-A) (A=37.5) Multiple of Deviation and Frequency (fd) 20-25 25-30 30-35 35-40 40-45 45-50 50-55 22.5 27.5 32.5 37.5 42.5 47.5 52.5 10 12 8 20 11 4 5 -15 -10 -5 0 5 10 15 -150 -120 -40 0 55 40 75 ∑f=70 ∑fd=140

=

$$\begin{array}{l}\bar{X}=\, \frac{\sum fd}{\sum f}\end{array}$$

=

$$\begin{array}{l}37.5\, +\, \frac{-140}{70}\end{array}$$

= 37.5 – 2 = 35.5

Question 5

Calculate the weighted mean from the following data

 Marks 60 75 63 59 55 Weight 2 1 5 5 3

Solution

 Marks (X) Weight (W) WX 60 75 63 59 55 2 1 5 5 3 120 75 315 295 165 ∑W=16 ∑WX=970
$$\begin{array}{l}\bar{X}_{w}\, =\, \frac{\sum WX}{\sum W}\end{array}$$

=

$$\begin{array}{l}\frac{970}{16}\end{array}$$
= 60.625

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