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 |
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 =
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,
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 =
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 |
=
=
= 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 |
=
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