Statistics MCQ Quiz - Objective Question with Answer for Statistics - Download Free PDF

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

∴ 13^th term = (416 + 507) - 900 = 923 - 900 = 23

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

⇒ 4^th + 5^th term

⇒ [(a – 2 + a – 0.5)/2]

⇒ [(2a – 2.5)/2]

⇒ a – 1.25

∴ The median of the data set is a – 1.25

Concept:

Variance (X) = \(\frac{\sum(X_i-\bar X)^2}{n}\) 

Standard Deviation = \(\sqrt{Variance }\)

Coefficient of variation for X = \(\frac{SD}{\bar X}\times 100 \) where X̅ is the means of observation.

Calculation:

For the prices of shares X:

Here, n = 10

X̅ = \(\frac{1}{n}\sum X_i\)

⇒ X̅ = \(\frac{35+54+52+53+56+58+52+50+51+49}{10}\) = \(\frac{480}{10}=48\)

Variance (X) = \(\frac{\sum(X_i-\bar X)^2}{n}\) = \(\frac{\sum d^2}{n}\) = \(\frac{980}{10}=98\)

SD(X) = \(\sqrt{Var(X)} \) = √98 = 9.9

Coefficient of variation for X = \(\frac{SD}{\bar X}\times 100 \) = \(\frac{9.9}{48}\times 100 = 20.6\)      ------(i)

For the prices of shares Y:

Here, n = 10

Y̅  = \(\frac{1}{n}\sum Y_i\)

⇒ Y̅  = \(\frac{108+107+105+105+106+107+104+103+104+101}{10}\) = \(\frac{1050}{10}=105\)

Variance (Y) = \(\frac{\sum(Y_i-\bar Y)^2}{n}\) = \(\frac{\sum di^2}{n}\) = \(\frac{40}{10}=4\)

SD(Y) = \(\sqrt{Var(Y)} \) = = √4 = 2

Coefficient of variation for Y = \(\frac{SD}{\bar Y}\times 100 \) = \(\frac{2}{105}\times 100 = 1.90 \)      ------(ii)

Comparing the equation (i) and (ii), coefficient of variation of Y is smaller than that of X.

∴ X is more stable than Y.

Concept:

Variance (X) = \(\frac{\sum(X_i-\bar X)^2}{n}\) 

Standard Deviation = \(\sqrt{Variance }\)

Coefficient of variation for X = \(\frac{SD}{\bar X}\times 100 \) where X̅ is the means of observation.

Calculation:

Here, n = 10

For the prices of shares Y:

Y̅  = \(\frac{1}{n}\sum Y_i\)

⇒ Y̅  = \(\frac{108+107+105+105+106+107+104+103+104+101}{10}\) = \(\frac{1050}{10}=105\)

Variance (Y) = \(\frac{\sum(Y_i-\bar Y)^2}{n}\) = \(\frac{\sum di^2}{n}\) = \(\frac{40}{10}=4\)

SD(Y) = \(\sqrt{Var(Y)} \)

⇒ SD(Y) = √4 = 2

∴ Standard deviation of Y is 2.

Concept:

Variance (X) = \(\frac{\sum(X_i-\bar X)^2}{n}\) 

Standard Deviation = \(\sqrt{Variance }\)

Coefficient of variation for X = \(\frac{SD}{̅ X}\times 100 \) where X̅ is the means of observation.

Calculation:

Here, n = 10

For the prices of shares X:

⇒ X̅ = \(\frac{1}{n}\sum X_i\)

⇒ X̅ = \(\frac{35+24+52+53+56+58+52+50+51+49}{10}\) = \(\frac{480}{10}=48\)

Variance (X) = \(\frac{\sum(X_i-\bar X)^2}{n}\) = \(\frac{\sum d^2}{n}\) = \(\frac{980}{10}=98\)

∴ The Variance of X is 98.

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

Formula used:

The mean of grouped data is given by,

\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)

Where, \(u_i \ = \ \frac{X_i\ -\ a}{h}\)

Xi = mean of ith class

fi = frequency corresponding to ith class

Given:

Calculation:

Now, to calculate the mean of data will have to find ∑fiXi and ∑fi as below,

Then,

We know that, mean of grouped data is given by

\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)

= \(\frac{1535}{43}\)

= 35.7

Hence, the mean of the grouped data is 35.7

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.

Formula Used∶

• σ² = ∑(x_i – x)²/n
• Standard deviation is same when each element is increased by the same constant

Calculation:

Since each data increases by 10,

There will be no change in standard deviation because (x_i – x) remains same.

∴ The standard deviation of 10, 11, 12, 13 _____ 19 will be will be K.

Alternate Method 

Concept:

Median: The median is the middle number in a sorted- ascending or descending list of numbers.

Case 1: If the number of observations (n) is even

\({\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}\)

Case 2: If the number of observations (n) is odd

\({\rm{Median\;}} = {\rm{value\;of\;}}{\left( {\frac{{{\rm{n}} + 1}}{2}} \right)^{{\rm{th}}}}{\rm{\;observation}}\)

Calculation:

Given values 2, 6, 6, 8, 4, 2, 7, 9

Arrange the observations in ascending order:

2, 2, 4, 6, 6, 7, 8, 9

Here, n = 8 = even

As we know, If n is even then,

\({\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}\)

= \(\rm \frac{4^{th} \;\text{observation}+5^{th} \;\text{observation}}{2} \)

= \(\frac{6+6}{2} =6\)

Hence Median = 6