Measures of Central Tendency: 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:

A firm buys machinery worth Rs. 2,00,000 every year and writes off depreciation as given below:

Find the average depreciation rate per year.

Solution:

Now, arranging the data for 3 parts, we find

Calculation average depreciation rate:

X̅w = ∑WX/∑W= 14200000/2000000 =7.1%

Problem 2:

The number 3.2, 5.8, 7.9 and 4.5 have frequencies x, (x + 2), (x – 3) and (x +6), respectively. If 4.876, find the value of x:

Solution:

Problem 3:

The average salary of male employees in a firm was Rs. 520 and that of female was Rs. 420; the mean salary of all the employees was Rs. 500.

Find the percentage of male and female employees:

Solution:

We represent by N₁and N₂ the number of male and female employees in the concern and X, and X, represent, respectively, their average salary. We denote X₁₂ as the average salary of all the workers in the firm.

Putting the all value, we get;

Thus, the percentage of male employees in the firm is

4/4+1×100 = 4/5 × 100 = 80%

Percentage of female = 100% – percentage of male

100% – 80% = 20%.

Problem 4:

The pass result of 40 students who took up a class test is given below:

If the average marks for all 50 students were 5.16, find out the average marks of the students who failed.

Solution:

Here,

Marks obtained by 40 students who passed:

Total marks of all 50 students = 50 x 5.16 = 258

Total marks of 40 students who passed = 237

∴ Marks of the remaining 10 students = 258 – 237 = 21

Hence, the average marks of 10 students

who failed = 21/10= 2.1 Mark.

Problem 5:

Calculate median from following data:

Solution:

Here, convert it first into exclusive form to get proper class-intervals and then proceed.

Problem 6:

Calculate the median for the following frequency distribution:

Solution:

First, arrange the data in ascending order and then find out median.

Calculation of median;

Problem 7:

The following is the table which gives you the distribution of marks secured by some students in the examination:

Find (i) Median marks (ii) The percentage of failure if the minimum for a pass is 35 marks.

Solution:

We are to make the class-boundaries from the class-limits given and then to find the cumulative frequency.

Median = Size of N/2th item

Solution:

The mark 35 represents the interval (34.5-35.5) taking marks as a continuous variable. Minimum pass marks is 34.5; number of students (f) obtaining less than 34.5 is the cumulative frequency corresponding to 34.5 marks. Now, using the above median formula, 34.5 = 30.5 + 4

Problem 8:

Find Median:

Solution:

Problem 9:

Compute median from the following data:

Solution:

Since we are given the mid-values, we should find out the upper and lower limits of the various classes.

Median = Size of N/2th item = size of 390/2 = 195th item.

Median lies in the class 150-160

Problem 10:

10% of the workers in a firm employing a total of 1,000 workers earn less than Rs. 5 per day, 200 earn between Rs. 5 and 9.99, 30% between Rs. 5 and 9.99, 30% between 10 and 14.99, 250 workers between 15 and 19.99 the rest Rs. 20 and the above.

What is the median wages?

Solution:

First, convert the given figure in a frequency distribution as follows:

Median = Size of N/2the item = Size of 1000/2= 500th item.

Median lies the class interval 10-14.99 but the real limit of this class 9.995-14.995.

Problem 11:

The following table shows the age distribution:

Find: (a) median (b) why is the median a more suitable measure of central tendency than the mean in the case?

Solution:

Firstly, the series is converted into continuous series,

(b) In the case of this distribution, median is more appropriate because the first class and last class, that is below 10 and 70 and above, respectively, are open end classes, AM is not appropriate but median is more appropriate.

Problem 12:

Grouping Method in Discrete Series (When question is silent):

Find the value of mode:

Solution:

In the question, the highest value of frequency is 10 and the succeeding value of that frequency is 9; if the difference is < 3 then we apply grouping method.

Analysis table

The highest frequency repeated is 6. Therefore, corresponding value 8 is mode. Note: If value is more than one time, highest values are repeated then it means, that is a biomodel series, the formula of bi-model series is,

Mode = 3 Median – 2 Mean

This is called empirical formula.

Problem 13:

Calculate the median and mode (by grouping method) from the following data:

Solution:

Firstly, central size (mid-point) converted into C.I.

Median = size of N/2th item 94/2= 47th item

Median lies in the class 40-50

Analysis table:

This series is bi-model series and hence we shall determine the mode by the indirect method Mode = 3 Mean – 2 Mean

Problem 14:

Calculate the value of mode by usual formula:

Solution:

Thus, mode lies between 60-70

Note:

It may be noted that in the above question the formula

Problem 15:

In the following wage distribution, the median and mode are Rs. 33.5 and Rs. 34, respectively, but three class frequencies are missing. Find out the missing frequency:

Let the missing frequencies be x, y, z as recorded in the table.

Now, N = x+ y + z + 30= 230(given)

∴ x + y + z = 230-30 = 200 …… (i)

Median is the value of Nth/2 = 230th/2 = 115th item

Median = 33.5 (given)

It lies in the class interval 30-40

Now, mode is 34, lies between 30-40

12y – 800 = 20y – 10y

or, 12y- 10y+ 10x = 800

or, 2y+ 10x = 800 …(iii)

Subtracting eqn. (iii) from (ii), we get

1.5y = 150

y = 150/1.5 = 100

Putting the value of y = 100 in eqn. (ii), we get

3.5 (100) + 10x = 950

350 + 10x = 950

10x = 950 – 350

10x = 600

x = 60

Putting the values of y = 100 and x = 60 in eqn. (i), we get

60+ 100 + z = 200

160 + z = 200

z = 200 – 160  =  40

x = 60, y = 100, z = 40

Problem 16:

Calculate the lower and upper Quartile. Third Deciles and Percentile from the following data:

Solution:

Since we are given mid-point, we will first find the lower and upper limits of the various classes. The method for finding out these limits is to take the difference between the two central values, divide it by 2, deduct the values so obtained from lower limits and add it to upper limits in the given case The first class shall be 0-5, 5-10 and so on.

Problem 17:

Find percentage of the students of a class getting marks (i) above 46 (ii) between 26 and 66 (iii) if 60% students pass this test, find minimum marks prescribed to pass the test.

Solution:

(i) No. of students above 46 marks = 50-46/10×70+ 55 + 35 +30

= 28 + 55 + 35 + 30 = 148

= % age 148 × 100/500 = 29.6

(ii) In this case minimum marks to pass the test are X, then (100-160) = 40% students get marks below X. So, if we find P₄₀, it should be equal to x.

Now, N₁ for P₄₀ (.N/100) = 40 × 500 = N₁= 200

P₄₀ = L₁+ N₁-c.f/f ×1

= 30 + 200-150/160 × 1 = 33.31 or 33

The minimum marks to pass the test are 33.

Problem 18:

Following is the distribution of marks obtained by 50 students in percentile law:

Calculate the median marks, if 60% of the students pass their test, find the minimum marks obtained by a pass candidate.

Solution:

The above data can be written in the following form:

Since, N/2 = 50/2 = 25, the median class is 20 – 30.

Therefore, Median = L₁ + N/2 –cf/f ×i

= 20 + 25 – 10/20 × 10 = 20 + 150/20 = 27.5

If 60% students pass, in order to find out the minimum marks obtained by a pass candidate, we have to calculate 40th percentile of the above distribution.

Since, 40(N/100) = 40×50/100=20, P₄₀ lies in the class 20-30

Hence P₄₀ = L₁+ 40N-c.f/f × i

= 20 + 20-10/20 × 10

= 20 + 100/20 = 20 + 5 = 25

Hence, minimum marks obtained by a pass candidate is 25.