Question

The marks of some students were listed out of 75. The standard deviation of marks was found to be 9. Subsequently, the marks were raised to a maximum of 100 and the variance of new marks was calculated. The new variance is:
(a) 144
(b) 122
(c) 81
(d) 3

Answer 1

Hint: Initially, the students’ marks are out of 75 and their standard deviation is 9. Now, we have raised the maximum marks to 100. Let us assume that initially, a student obtains x marks out of 75. Now, by the unitary method, we will find that student’s marks out of 100. Then each observation is changed by some factor. This means our standard deviation is also raised by that factor. Multiply that factor to get the new standard deviation and square this new standard deviation to get the variance.

Complete step-by-step solution:
When the students' marks are listed out of 75 then the standard deviation is 9. After that, the maximum marks are raised to 100 which will automatically change the marks of the students.
Now, let us assume that a student obtains x marks out of 75 so we can say that out of 75, a student gets x marks. Now, the marks that student got out of 1 mark is equal to:
$\dfrac{x}{75}$
And the marks that student got out of 100 marks is calculated by multiplying 100 to the above expression which will make the above expression as:
$\dfrac{x}{75}\times 100$
In the above expression, the numerator and the denominator can be divided by 25 and we get,
$\dfrac{4x}{3}$
From the above, we got that out of 100 that student marks become $\dfrac{4x}{3}$. Due to the change in marks of the students, the standard deviation will also change so now the standard deviation becomes $\dfrac{4}{3}$ of initial standard deviation because each observation will get multiplied by $\dfrac{4}{3}$.
New standard deviation is equal to:
$\begin{align}
  & \dfrac{4}{3}\times 9 \\
 & =4\times 3=12 \\
\end{align}$
We know that variance is the square of the standard deviation so squaring 12 we get,
$\begin{align}
  & {{\left( 12 \right)}^{2}} \\
 & =144 \\
\end{align}$
Hence, the new variance is equal to 144 and the correct option is (a).

Note: You might be thinking that if each observation gets multiplied by $\dfrac{4}{3}$ then why do we have multiplied standard deviation by $\dfrac{4}{3}$. The answer is as follows:
The formula for standard deviation is:
\[\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}^{2}}}{n}-{{\left( \dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n} \right)}^{2}}}\]
Now, if each observation gets multiplied by $\dfrac{4}{3}$ the above expression will look like:
\[\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( \dfrac{4}{3}{{x}_{i}} \right)}^{2}}}}{n}-{{\left( \dfrac{\sum\limits_{i=1}^{n}{\dfrac{4}{3}{{x}_{i}}}}{n} \right)}^{2}}}\]
\[=\sqrt{{{\left( \dfrac{4}{3} \right)}^{2}}\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}^{2}}}{n}-{{\left( \dfrac{4}{3} \right)}^{2}}{{\left( \dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n} \right)}^{2}}}\]
Now, taking ${{\left( \dfrac{4}{3} \right)}^{2}}$ out from the square root then the above expression will look like:
\[\dfrac{4}{3}\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}^{2}}}{n}-{{\left( \dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n} \right)}^{2}}}\]
As you can see that our standard deviation got multiplied by $\dfrac{4}{3}$.