Question

A group of \[100\] candidates have their average height $163.8$cm with coefficient of variation $3.2$. What is the standard deviation of their heights?
A.$5.24$
B.$2.24$
C.$7.24$
D.None of these

Answer 1

Hint: We are given coefficient of variation, average of the heights of students and we have to find the standard deviation of their heights. We will use the formula which gives the relation between coefficients of variation, standard deviation and mean which is given as-
$ \Rightarrow $ C.V. =$\dfrac{\sigma }{{\bar x}} \times 100$ where $\bar x \ne 0$
Where C.V. is coefficient of variation, $\sigma $ is the standard deviation and $\bar x$ is the mean or average
Now, we will put the given values in the formula and solve it to get the standard deviation of the heights of students.

Complete step-by-step answer:
Given, the total number of candidates =\[100\]
Coefficient variation is =$3.2$
And the mean of their heights is =$163.8$cm
We have to find the standard deviation of their heights.
Now, we know that coefficient of variation is related to standard deviation and mean as follows-
$ \Rightarrow $ C.V. =$\dfrac{\sigma }{{\bar x}} \times 100$ where $\bar x \ne 0$
Where C.V. is coefficient of variation, $\sigma $ is the standard deviation and $\bar x$ is the mean or average
On putting C.V. =$3.2$, $\bar x$=$163.8$cm, we get-
$ \Rightarrow 3.2 = \dfrac{\sigma }{{163.8}} \times 100$
On rearranging, we get-
$ \Rightarrow \sigma = \dfrac{{3.2}}{{100}} \times 163.8$
On multiplication, we get-
$ \Rightarrow \sigma = \dfrac{{524.16}}{{100}}$
On division, we get-
$ \Rightarrow \sigma = 5.2416$
Hence the correct answer is option A.

Note: Standard deviation is the method of calculating the measure of variations. It is the positive square root of a variance of a variate. We use the formula given in the solution to compare the variability of two series with data expressed in different units.