Statistics Test - 53

Given:

Mean = 5 and Standard deviation = 2

If 5 is added to each value,

We know that adding a constant to each value Mean will be changed by adding a constant value and standard deviation remain the same.

Therefore, Mean = 5 + 5 = 10

Standard deviation will not change.

Now,

Coefficient of variation \(=\frac{\text { standard deviaiton }}{\text { Mean }} \times 100=\frac{2}{10} \times 100\)

∴ Coefficient of variation = 20

We know that:

Median \(=\left[\frac{(n+1) }{2}\right]^{\text {th }}\) term

\(n \rightarrow\) odd term 

Median \(=\left[\frac{(5+1)}{2}\right]^{\text {th }}\) term

Median \(=3^{\text {th }}\) term

\(\therefore\) Median of \(1,2,3,4\) and \(5\) is \(3\).

Given:

Arithmetic mean \(=(\mu)=10\)

Coefficient of variation \(= CV =40 \%\) or \(0.4\)

We know that:

\(C V=\frac{\sigma}{\mu}\)

\(\sigma=\) Standard deviation

\(\mu=\) Mean

\(0.4=\frac{\sigma}{10}\)

\(\Rightarrow \sigma=\) (Standard deviation \(=0.4 \times 10=4\)

Now, Variance \(=V=\) standard deviation) \(^{2}=\sigma^{2}\)

\(\Rightarrow\) Variance \(=4^{2}=16\)

We know that:

\(V(a x+b)=a^{2}(V(x)\)

\(\Rightarrow V(10-2 x)=(-2)^{2} V(X)\)

\(\Rightarrow V(10-2 x)=4 V(x)\)

\(\therefore V(10-2 x)=4 \times 16=64\)

Given:

Given data is 3, 5, 8, 6, 9, 4, 5, 2, 7, 1, 5

We know that:

Mean of given observations \(= \frac{(\text{Sum of all the observations})}{(\text{Number of observations})}\)

\(\frac{(3 + 5 + 8 + 6 + 9 + 4 + 5 + 2 + 7 + 1 + 5)}{11}\)

\(= 5\)

∴ The mean of the given data is 5.

n observations are given by:

\(a_{1}, a_{2}, a_{3}, \ldots ..a_{m}+...a_{n}\)

Mean of "m" observation is "\(n\)"

\( \frac{a_{1}+a_{2}+a_{3} \ldots \ldots a_{m}}{m}=n \quad \quad \ldots\)(1)

Mean of remaining observation is \(m\) so number of such observation are \(n -3\)

\(\text { Mean }=\frac{a_{m+1}+\ldots+a_{n}}{n-m}=m \quad \quad \ldots\)(2)

By rearranging & adding equation (1) & (2)

\(a_{1}+a_{2}+\ldots \ldots a_{m+1} \ldots \ldots+a_{n}=n m+m(n-m)\)

\(=2 m n-m^{2}\)

So, total observation is "\(n\)"

So, mean is \(\frac{2 m n-m^{2}}{n}\)

\(=2 m-\frac{m^{2}}{n}\)

Here, we have to choose the option which will give the coefficient of variation as \(30 \%\)

Option (D): When mean is 10 and standard deviation is 3

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

\(\Rightarrow\) Coefficient of variation \(=\frac{3}{10} \times 100=30 \%\)

Option (A): When mean is 30 and standard deviation is 15

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

\(\Rightarrow\) Coefficient of variation \(=\frac{15}{30} \times 100=50 \%\)

Thus, option (A) is not the correct answer.

Option (B): When mean is 4 and standard deviation is 16

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

\(\Rightarrow\) Coefficient of variation \(=\frac{16}{4} \times 100=400 \%\)

Thus, option (B) is not the correct answer.

Option (C): When mean is 3 and standard deviation is 3

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

\(\Rightarrow\) Coefficient of variation \(=\frac{3}{3} \times 100=100 \%\)

The mean deviation is the sum of the deviations from the average divided by the number of items in the series.

\(MD =\left(\frac{1}{n}\right)\sum \mid x - M \mid\)

\(\therefore M\) is Mean or the median or mode.

Given: Mean of 100 observations is 50 and standard deviation is 10.

If 5 is added to each observation

∵ There are total 100 observations

500 is added to sum of the observations and now dividing it by the no. of observations, we get that the mean is increased by 5.

As we know that, the standard deviation of N observations is given by:

\(\sigma=\sqrt{\frac{1}{N} \times \sum_{i=1}^{N}\left(x_{i}-\mu\right)^{2}}\) where, \(\mu\) is the arithmetic mean

∵ Every observation increased by 5 and mean also increased by 5 and standard deviation is the square root of the difference of mean and each observation divided by the number of observations.

So, standard deviation will remain same.

Thus, new mean is 55 and new standard deviation is still 10.

Given:

The mean of 10 observations \(x_{1}, x_{2}, x_{3} \ldots . x_{10}\) is 20.

Mean of \(x_{1}+2, x_{2}+4, x_{3}+6, \ldots x_{10}+20=\) Mean of \(\left(x_{1}, x_{2}, x_{3} \ldots x_{10}\right)+\) Mean of \((2,4)\) \((6, \ldots .20) \ldots(1)\)

Now mean of \((2,4,6, \ldots .20)\) is:

Mean \(= \frac{\text{Sum of total observations}}{\text{Total number of observation}}\)

\(\frac{(2+4+6+8+10+12+14+16+18+20)}{10}\)

\(=\frac{2 \times 55}{10}=11\)

From equation (1),

Mean of \(x_{1}+2, x_{2}+4, x_{3}+6, \ldots x_{10}+20=20+11=31\)

Given: \(x=\) M.D., \(y=\) S.D

We know that,

M.D \(=\frac{4}{5}\) S.D

Where, M.D is mean deviation and S.D is standard deviation

\( x =\frac{4}{5} y\)

\(\therefore x < y\)