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

Given

Observations = 3, 6, 10

Formula used

Harmonic mean = n/Σ1/x_i where i = 1, 2, 3, 4 ------

n = number of observation

x = observations value

Calculation

x_i = 3, 6, 10

⇒ Σ1/xi  = 1/3 + 1/6 + 1/10 = 3/5

⇒ number of observation = 3

⇒ Harmonic mean  = 3/3/5 = (3 × 5)/3

∴ Harmonic mean = 5

Important Points

The harmonic mean it is defined as the reciprocal  of the arithmetic mean of the observations of a set

Harmonic mean = n/Σ1/xi where i = 1, 2, 3, 4 ------

n = number of observation

x = observations value

Geometric mean = it  is defined as the n-th root of the product of the n observations.

GM = n(√x₁ × x₂--------x_n)

Given

Regression line of y on x

4y = 15 + 6x

y = 15/4 + (6/4)x

Coefficient of correlation between x and y = ρ_xy = 0.9

Formula

If the regression coefficient of x on y be b_xy, then

ρ_xy = √(b_yx × b_xy)

Calculation

b_xy = 6/4 = 1.5

⇒ 0.9 =√(1.5 × b_xy)

⇒ b_xy = (0.9 × 0.9)/1.5

⇒ b_xy = 0.54

∴ The regression coefficient of x on y is 0.54 

Formula

Spearman’s rank correlation coefficient is denoted by ρ

ρ = 1 – [6∑d²/n(n² – 1)]

d = difference between rank of paied items  of x and y variables

∑d² = total of squares of rank differences

n = number of pairs of items

Given

∑d² = ∑(R₁ – R₂)² = 21

n = 6

Calculation

ρ = 1 – [6 × 21/6(6² – 1)

ρ = 1 – 21/35

ρ = 14/35

∴ The rank correlation coefficient is 0.4

Given

Coefficient of correlation r = 0.5

Formula

Coefficient of determination = r²

Coefficient of non determination = 1 - coefficient of determination

Calculation

r = 0.5

⇒ r² = 0.25

∴ The coefficient of determination = 0.25

∴ The coefficient of non determination is s1 - 0.25 = 0.75

Given

Arithmetic mean =  (μ) = 16

variance = σ² = 4

Formula

Standard deviation = σ = √variance

CV = Coefficient of variation = σ/μ

σ = Standard deviation

μ = Mean

Calculation

Standard deviation =  σ = √4 = 2

⇒ CV = (σ/μ) × 100

⇒ CV = (2/16) × 100

∴ Coefficient of variation is 12.5

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

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:

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.

Shortcut Trick

Use assumed mean (A) = 15.

Deviations (d = x − A) are −4, −2, 0, 2, 4 for the respective class mid-points.

Net deviation = 6 × (−4) + 8 × (−2) + 5 × 0 + 7 × 2 + 4 × 4 = −10.

Average deviation = −10 ÷ 30 = −0.33.

Actual Mean = 15 − 0.33 = 14.67.

∴ The correct answer is 14.67.

Alternate Method

Given:

Class intervals and corresponding frequencies representing marks of students.

Formula Used:

Mean = Σ(f × x) ÷ Σf

Where x = Mid-point of class interval = (Lower limit + Upper limit) ÷ 2

Calculations:

⇒ For 10 − 12: x = 11, f = 6 → f × x = 66

⇒ For 12 − 14: x = 13, f = 8 → f × x = 104

⇒ For 14 − 16: x = 15, f = 5 → f × x = 75

⇒ For 16 − 18: x = 17, f = 7 → f × x = 119

⇒ For 18 − 20: x = 19, f = 4 → f × x = 76

⇒ Total Frequency (Σf) = 6 + 8 + 5 + 7 + 4 = 30

⇒ Total Sum Σ(f × x) = 66 + 104 + 75 + 119 + 76 = 440

⇒ Mean = 440 ÷ 30 = 44 ÷ 3

⇒ Mean = 14.666...

⇒ Rounding to two decimal places = 14.67

∴ The correct answer is 14.67.

Additional Information

Direct Mean Method

Mean = Σ(f × x) ÷ Σf, where x is the mid-point of the class interval and f is the frequency. It is best used when numerical values of x and f are small.

Assumed Mean Method

Mean = A + [Σ(f × d) ÷ Σf], where A is the assumed mean and d = x − A. This heavily simplifies calculations when values of x and f are large.

Step-Deviation Method

Mean = A + h × [Σ(f × u) ÷ Σf], where u = (x − A) ÷ h and h is the common class width. This further reduces calculation complexity for grouped distributions.

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

Calculation:

Let the numbers be x_1, x₂, x₃, x₄.

The mean of four numbers x_1, x₂, x₃, x₄ = 37

The sum of four numbers x_1, x₂, x₃, x₄ = 37 × 4 = 148.

The mean of the smallest three numbers x_1, x₂, x₃ = 34

The sum of the smallest three numbers x_1, x₂, x₃ = 34 × 3 = 102.

∴ The value of the largest number x₄ = 148 – 102 = 46.

The Range (Difference between largest and smallest value) x₄ – x₁ = 15.

∴ Smallest number x₁ = 46 – 15 = 31.

Now,

The sum of x₂, x₃ = Total sum – (sum of smallest and largest number).

⇒ 148 – (46 + 31)

⇒ 148 – 77

⇒ 71

Now,

Concept:

Standard deviation:

The standard deviation of the observation set \(\rm \{x_i,i=1,2,3,\cdots\}\) is given as follows:

\(\rm \sigma=\sqrt{\dfrac{\sum\left(x_i-\mu\right)^2}{N}}\)

Where \(\rm N=\mbox{size of the observation set}\) and \(\rm \mu=\mbox{mean of the observations}\) .

Calculations:

First, we will calculate the mean of the given observations.

\(\begin{align*} \mu &= \dfrac{-\sqrt6-\sqrt5-\sqrt4-1+1+\sqrt4+\sqrt5+\sqrt6}{8}= 0 \end{align*}\)

Therefore, the numerator inside the square root term of the standard deviation formula will simply be equal to \(\rm (x_i-\mu)^2=x_i^2\) .

Now we observe that \(\rm N=8\) .

Therefore, the standard deviation is given as follows:

\(\begin{align*} \sigma &= \sqrt{\dfrac{\left(-\sqrt6\right)^2+\left(-\sqrt5\right)^2+\left(-\sqrt4\right)^2+\left(-1\right)^2+\left(1\right)^2+\left(\sqrt4\right)^2+\left(\sqrt5\right)^2+\left(\sqrt6\right)^2}{8}}\\ &= \sqrt{\dfrac{32}{8}}\\ &= \sqrt4\\ &= 2 \end{align*}\)

Therefore, the standard deviation of the given observations is 2.

Concept:

• \({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)

Calculation:

We know that, \({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)

\(\therefore {\rm{Mean}} = \frac{{800}}{{24}} = 33.3\)

Concept:

Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)

Variance = (Standard Deviation)²

Calculation:

Given coefficient of variation = 45% = 0.45

And mean = 100

As Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)

0.45 = \(\rm\text{Standard Deviation} \over100\)

Standard Deviation = 100 × 0.45

SD = 45

∴ Variance = 45² = 2025