Measures of Central Tendency: 10 Problems and Solutions | Statistics

Here is a compilation of top twenty-eight problems on the measures of central tendency along with its relevant solutions.

Problem 1:

Calculate the mean using different methods:

Solution:

(i) X̅ Under direct method

X̅ = 3300/100 = 33

(ii) X̅ short-cut method

X̅ = 25 + 800/100 = 25 + 8 = 33 100

(iii) X̅ under step deviation method

X̅= 25 + 80/100 x 10 = 25 + 8 = 33.

Problem 2:

Find mean of the following data:

Solution:

In the above statement, the difference between two variables are differ by 10, 20, the last difference between two variables is 60 because the previous difference is 50.

Note:

It is regarded essential to convert all other types of series into exclusive type: otherwise we will proceed to a wrong result.

Problem 3:

Find mean.

Solution:

The difference between two variables in class interval that is 10, so now

Problem 4:

Solution:

Problem 5:

From the following distribution of marks obtained by 50 students in quantitative methods. Calculate Arithmetic Mean:

Solution:

The data is in cumulative form. Let us first convert it into simple frequency distribution. It is given that 50 students have marks over 10 and 46 over 20, the number of students having marks between 10-20 is 50 – 46 = 4. In the same manner, frequencies of 20 – 30, 30 – 40 can be computed.

Problem 6:

The arithmetic mean of 50 terms was calculated as 30.2, but later on it was known that without increasing the number of terms, one term with a value of 36 was included twice. Find the correct mean.

Solution:

Problem 7:

X̅ of 40 terms is 30 but two terms 13 and 24 were misread as 22 and 48. Find correct X̅ if:

(a) It is an individual series

(b) It is continuous series as:

(i) 10-15, 15-20, 20-25

(ii) 10-20, 20-30, 30-40….

Solution:

X̅ of 40 terms = 30

Total of 40 terms (∑X) = 40 x 30 = 1200

(a) and (b); it is an individual or discrete series case

(Incorrect) total = 1200

Correct total = 1200 + 13 + 24 – 22 – 48 = 1167

Correct X̅= 1167/40 = 29.175

(b) If it is continuous series

(i) With class intervals 10-15, 15-20, 20-25

Correct total = 1200 + 12.5 + 22.5 – 22.5 – 47.5

= 1235 -70= 1165

Correct X̅= 1167/40= 29.125

(ii) With class intervals 10-20, 20-30, 30-40

Correct total = 1200 + 15 + 25 – 25 – 45 = 1170

Correct X̅= 1170/40= 29.25 40

Problem 8:

The mean of the following series is 30.5, find missing figure.

Problem 9:

The following are the monthly salaries in rupees of 20 employees in a firm:

The firm gave bonus of Rs.10, 15, 20, 25 and 30 to the individual in the respective salary groups exceeding 60 but not exceeding Rs.80, exceeding Rs.80 but not exceeding Rs.100, and so on up to exceeding Rs.140 but not exceeding Rs160.

Find:

(i) The total bonus paid.

(ii) Average bonus paid per employee.

Solution:

∴ Total bonus paid = ∑fx = Rs =390

Average Bonus paid X̅ = ∑fx/N = 390/20 = 19.5

Problem 10:

Calculate the class interval if mean = 33 and assumed mean = 35:

Solution:

I^st class intervals 5-5 to 5 + 5 = 0-10

II^nd class intervals 15 – 5 to 15 + 5 = 10-20

III^rd class intervals 25 – 5 to 25 + 5 = 20-30

IV^th class intervals 35 – 5 to 35 + 5 = 30-40

V^th class intervals 45 – 5 to 45 + 5 = 40-50

VI^th class intervals 55 – 5 to 55 + 5 = 50-60