Test: Measures of Central Tendency - Civil Engineering (CE) MCQ

For a distribution, Median

where, L = lower class limit of median class
N = Sum of frequencies
CF = cumulative frequency of class preceding the median class
f = frequency of median class
h = class length of median class.

∑f = N = 20
⇒ N/2 = 10 
As 7 < 10 < 13, N/2 lies in the class length 30 - 40 by comparing cumulative frequency.
Median class = 30 - 40
⇒ L = 30, N = 60, CF = 7, f = 6, h = 40 - 30 - 10

∴ The correct option is (3).

Given:
The mean of 25 observations is 36
The mean of the first 13 observations is 32 and that of the last 13 observations is 39 

Concept used:
Mean = sum of all observation/total number of observation

Calculation:
The sum of all 25 observation = 25 × 36 = 900
Sum of first 13 observations = 13 × 32 = 416
Sum of last 13 observations = 13 × 39 = 507
∴ 13th term = (416 + 507) - 900 = 923 - 900 = 23

Given:
The given data is 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Concept used:
The mode is the value that appears most frequently in a data set
At the time of finding Median
First, arrange the given data in the ascending order and then find the term

Formula used:
Mean = Sum of all the terms/Total number of terms
Median = {(n + 1)/2}th term when n is odd 
Median = 1/2[(n/2)th term + {(n/2) + 1}th] term when n is even
Range = Maximum value – Minimum value 

Calculation:
Arranging the given data in ascending order 
2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19
Here, Most frequent data is 4 so 
Mode = 4
Total terms in the given data, (n) = 15 (It is odd)
Median = {(n + 1)/2}th term when n is odd 
⇒ {(15 + 1)/2}th term 
⇒ (8)th term
⇒ 6 
Now, Range = Maximum value – Minimum value 
⇒ 19 – 2 = 17
Mean of Range, Mode and median = (Range + Mode + Median)/3
⇒ (17 + 4 + 6)/3 
⇒ 27/3 = 9
∴ The mean of the Range, Mode and Median is 9

Concept:
Mean: The mean or average of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set.
Mode: The mode is the value that appears most frequently in a data set.
Median: The median is a numeric value that separates the higher half of a set from the lower half. 
Relation b/w mean, mode and median:
Mode = 3(Median) - 2(Mean)

Calculation:
Given that,
mean of data = 4 and mode of  data = 10
We know that
Mode = 3(Median) - 2(Mean)
⇒ 10 = 3(median) - 2(4)
⇒ 3(median) = 18
⇒ median = 6
Hence, the median of data will be 6.

Concept:

To find number of observations between 250 and 300.
first we have to draw a frequency distribution table from this data.

∴ The Number of observation in between 250-300 = 38 - 15 = 23.

Concept:

Calculation:

∑f = N = 20
⇒ N/2 = 10 
As 7 < 10 < 13, N/2 lies in the class length 30 - 40 by comparing cumulative frequency.
Median class = 30 - 40
⇒ L = 30, N = 60, CF = 7, f = 6, h = 40 - 30 - 10

Now calculating mean deviation about median,

⇒ Mean deviation about median 

∴ The correct option is (4).

Concept:

Calculation:
Given: x is the mean of squares of first n natural numbers and y is the square of mean of first n natural numbers



⇒ 2 x 42(2n + 1) = 3 x 55(n + 1)
⇒ 168n + 84 = 165n + 165
⇒ 3n = 81
⇒ n = 27
∴ The correct option is (3).

Given:
The given values =  a + 4, a – 3.5, a – 2.5, a – 3, a – 2, a + 0.5, a + 5 and a – 0.5

Concept used:
If n is odd
Median = [(n + 1)/2]th observations
If n is even
Median = [(n/2)th + (n/2 + 1)th observations]/2
Calculation:
a + 4, a – 3.5, a – 2.5, a – 3, a – 2, a + 0.5, a + 5 and a – 0.5
Arrange the data in ascending order
⇒ a – 3.5, a – 3, a – 2.5, a – 2, a – 0.5, a + 0.5, a + 4, a + 5
Here, the n is 8, which is even
Median =  [(n/2)th + (n/2 + 1)th observations]/2
⇒ [(8/2) + (8/2 + 1)/2] term
⇒ 4th + 5th term
⇒ [(a – 2 + a – 0.5)/2]
⇒ [(2a – 2.5)/2]
⇒ a – 1.25
∴ The median of the data set is a – 1.25

Formula used:

The mean of grouped data is given by,

X_i = mean of ith class
f_i = frequency corresponding to ith class
Given:

Calculation:
Now, to calculate the mean of data will have to find ∑f_iX_i and ∑f_i as below,

Then,
We know that, mean of grouped data is given by

Hence, the mean of the grouped data is 35.7