Variance and Standard Deviation MCQ Quiz - Objective Question with Answer for Variance and Standard Deviation - Download Free PDF

Explanation:

Given:

\(\rm \Sigma_{i=1}^9x_i^2=885\)

Variance σ = \(\rm \Sigma_{i=1}^9x_i^2 - (mean)^2\)

= \(\frac{855}{9} -(M)^2\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)⇒ σ + M² = \(\frac{855}{9} = 95\)

∴ Option (b) is correct.

Calculation

Let the two unknown items be x and y. Then,

Mean = 4 ⇒ \(\frac{1+2+6+x+y}{5}=4\)

⇒ x + y = 11...(i)

and variance = 5.2

⇒ \(\frac{1^{2}+2^{2}+6^{2}+x^{2}+y^{2}}{5}\) - (mean)2 = 5.2

⇒ 41 + x2 + y2 = 5(5.2 + 16)

⇒ 41 + x2 + y2 = 106

⇒ x2 + y2 = 65... (ii)

On solving Eqs. (i) and (ii), we get

x = 4, y = 7 or x = 7, y = 4

sum = 28

Concept Used:

Mean: \(\bar{x} = \frac{\sum x_i}{n}\)

Variance: \(\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2\)

Calculation:

Given:

Mean of 5 observations = 4

Variance of 5 observations = 5.2

Three observations: 1, 2, 6

Let the other two observations be x and y.

Mean: \(\frac{1 + 2 + 6 + x + y}{5} = 4\)

⇒ \(9 + x + y = 20\)

⇒ \(x + y = 11\) ...(1)

Variance: \(\frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - 4^2 = 5.2\)

⇒ \(\frac{1 + 4 + 36 + x^2 + y^2}{5} - 16 = 5.2\)

⇒ \(\frac{41 + x^2 + y^2}{5} = 21.2\)

⇒ \(41 + x^2 + y^2 = 106\)

⇒ \(x^2 + y^2 = 65\) ...(2)

From (1), \(y = 11 - x\)

Substitute in (2):

⇒ \(x^2 + (11 - x)^2 = 65\)

⇒ \(x^2 + 121 - 22x + x^2 = 65\)

⇒ \(2x^2 - 22x + 56 = 0\)

⇒ \(x^2 - 11x + 28 = 0\)

⇒ \((x - 4)(x - 7) = 0\)

⇒ \(x = 4\) or \(x = 7\)

If \(x = 4\), \(y = 11 - 4 = 7\)

If \(x = 7\), \(y = 11 - 7 = 4\)

The other two observations are 4 and 7.

Hence option 1 is correct

Concept Used:

Coefficient of Skewness = \(\frac{Mean - Mode}{Standard Deviation}\)

Calculation

Given:

Coefficient of Skewness = 0.32

Standard Deviation (SD) = 6.5

Mean = 29.6

Substitute the given values into the formula:

⇒ \( 0.32 = \frac{29.6 - Mode}{6.5} \)

⇒ \( 0.32 \times 6.5 = 29.6 - Mode \)

⇒ \( 2.08 = 29.6 - Mode \)

⇒ \( Mode = 29.6 - 2.08 \)

⇒ \( Mode = 27.52 \)

∴ The mode of the distribution is 27.52.

Hence option 2 is correct

\(|A|=\left|\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right|\)

= a₁₁ a₂₂ – a₂₁

a₁₂ = {–1, 0, 1}

\(\begin{array}{c|c|c|c} \mathrm{x} & \mathrm{P}_{\mathrm{i}} & \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}} & \mathrm{P}_{1} \mathrm{X}_{\mathrm{i}}{ }^{2} \\ -1 & \frac{3}{16} & -\frac{3}{16} & \frac{3}{16} \\ 0 & \frac{10}{16} & 0 & 0 \\ 1 & \frac{3}{16} & \frac{3}{16} & \frac{3}{16} \\ \hline & & \sum \mathrm{P}_{\mathrm{i}} X_{\mathrm{i}}=0 & \sum \mathrm{P}_{\mathrm{i}} X_{\mathrm{i}}{ }^{2}=\frac{3}{8} \end{array}\)

\(\therefore \operatorname{var}(\mathrm{x})=\sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}^{2}-\left(\sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}\right)^{2} \)

\(=\frac{3}{8}-0=\frac{3}{8} \)

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 

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.

\(\bar x\left( {mean} \right)\; = \;\frac{{\sum fx}}{n}\)

⇒ n = total frequency

\(\sum fx = Sum\;of\;the\;product\;of\;mid - interval\;values\;and\;their\;corresponding\;frequency\;\;\)

Mid value of 10 – 12 = (10 + 12)/2 = 11

Mid value of 12 – 14 = (12 + 14 )/2 = 13

Mid value of 14 – 16 = (14 + 16 )/2 = 15

Mid value of 16 – 18 = (16 + 18 )/2 = 17

Mid value of 18 – 20 = (18 + 20 )/2 = 19

\(\Rightarrow \;Mean\; = \;\frac{{11\; \times \;6\; + \;13\; \times \;8\; + \;15\; \times \;5\; + \;17\; \times \;7\; + \;19\; \times \;4}}{{6\; + \;8\; + \;5\; + \;7\; + \;4}} = \;\frac{{440}}{{30}}\)

⇒ Mean = 14.67

∴The mean marks of the given data are 14.67

Concept:

If every observation is multiplied by a number, then the mean is also multiplied by the same number 

\(\text { Variance }=\sigma^{2}=\frac{\sum\left(x_{i}-\bar{x}\right)^{2}}{\text {} n}\)

If every observation is multiplied by a number, new variance = (number)² × old variance

Calculation:

Here, mean (x̅) = 4 and variance (σ²) = 2

Number of observation (n) = 10

New mean = 2 × (mean)

= 2 × 4

⇒ 8

New var = (number)2 × old variance

⇒ 2² × 2

⇒ 8

Hence, option (3) is correct.

Concept:

\({\rm{Coefficient\;of\;variation}} = {\rm{}}\frac{{{\rm{standard\;deviaiton}}}}{{{\rm{Mean}}}} \times 100\)

\(\rm S.D = \sqrt {Variance}\)

Calculation:

Given:

Variance = 81 and coefficient variation = 30%

We know that, \(\rm S.D = \sqrt {Variance}\)

\(\rm S.D = \sqrt {Variance} = \sqrt {81} = 9\)

As we know, \({\rm{Coefficient\;of\;variation}} = {\rm{}}\frac{{{\rm{standard\;deviaiton}}}}{{{\rm{Mean}}}} \times 100\)

\(\Rightarrow {\rm{Coefficient\;of\;variation}} = {\rm{}}\frac{9}{\rm Mean} \times 100 \)

\(\Rightarrow \rm 30 = \frac{9}{{Mean}} \times 100\)

⇒ Mean = 30

Calculation:

Given values 6.5, 3.4, 8.6, 2.9

By arranging the given in the ascending order, we get

2.9, 3.4, 6.5, 8.6

Explanation:

The terms used in calculating standard deviation are deviations from the mean of the observations.
As each number/observation is increased by 2, the deviations from the mean remain the same.
Hence standard deviation remains the same.
On the other hand, median gives the middle term or the average of the two middle terms accordingly when the total number of terms is odd or even.
Thus, it has to increase by 2. 

Hence new median = 20 + 2 = 22 and SD = 4

Mean, \(\bar x = \frac{{2 + 4 + 6 + 8 + 10}}{5} = \frac{{30}}{5} = 6\) 

\(\therefore Variance = \frac{1}{n}\sum {\left( {{x_i} - \bar x} \right)^2}\;\)

\(= \frac{1}{5}\left\{ {{{\left( {2 - 6} \right)}^2} + {{\left( {4 - 6} \right)}^2} + {{\left( {6 - 6} \right)}^2} + {{\left( {8 - 6} \right)}^2} + {{\left( {10 - 6} \right)}^2}} \right\}\)

\(= \frac{1}{5}\left\{ {16 + 4 + 0 + 4 + 16} \right\} = \frac{1}{5} \times 40 = 8\)

Z is the midpoint of range of x for every data

sum of all F × Z = 875 + 35P

mean of above data is 25.2 & sum of all F = 8 + 12 + 10 + P + 9 = 39 + P

mean \(= \frac{{875 + 35P}}{{39 + P}} = 25.2\)

⇒ 9.8 P = 107.8