Question

The scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane’s score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students score above Jane.
(A) 109
(B) 106
(C) 150
(D) 160

Answer 1

Hint: We start solving this question by first assuming the random variable $X$ denotes the scores of students. Then we assume the mean and standard deviation of the given distribution of scores are $\mu $ and $\sigma $. Then we write the score of Jane in terms of $\mu $ and $\sigma $. Then we find the value of $P\left( X>Jane's\ Score \right)$. Then we assume the random variable $Z$ such that $Z=\dfrac{X-\mu }{\sigma }$. Then we consider the property that, if $X$ follows normal distribution with mean $\mu $ and standard deviation $\sigma $, then $Z=\dfrac{X-\mu }{\sigma }$ follows standard normal distribution. Then we find the probability value required from the normal distribution table. Then we multiply the obtained probability with the total number of students to find the number of students that score above Jane.

Complete step by step answer:
We are given that scores of 500 students are normally distributed.
We are also given that Jane’s score is 0.8 standard deviation above mean.
So, let us assume that mean and standard deviation of the given distribution of scores are $\mu $ and $\sigma $.
Let us assume that the random variable $X$ denotes the scores of the students. So, it follows normal distribution with mean $\mu $ and standard deviation $\sigma $.
As we are given that Jane’s score is 0.8 standard deviation above mean, we can write it as,
$\Rightarrow Jane's\ Score=0.8\sigma +\mu $
We need to find the number of students whose score is above Jane.
First let us find the probability of a student scoring above Jane, that can be written as,
$\Rightarrow P\left( X>Jane's\ Score \right)$
Now let us substitute the score of Jane. Then we get the probability as,
$\Rightarrow P\left( X>0.8\sigma +\mu \right)$
We can write the above equation as,
$\begin{align}
  & \Rightarrow P\left( X>0.8\sigma +\mu \right)=P\left( X-\mu >0.8\sigma \right) \\
 & \Rightarrow P\left( X>0.8\sigma +\mu \right)=P\left( \dfrac{X-\mu }{\sigma }>0.8 \right).........\left( 1 \right) \\
\end{align}$
Now let us assume the random variable $Z$ such that $Z=\dfrac{X-\mu }{\sigma }$. Then we get,
$\Rightarrow P\left( \dfrac{X-\mu }{\sigma }>0.8 \right)=P\left( Z>0.8 \right)..........\left( 2 \right)$
Now let us consider the property, If $X$ follows normal distribution with mean $\mu $ and standard deviation $\sigma $, then $Z=\dfrac{X-\mu }{\sigma }$ follows standard normal distribution, that is normal distribution with mean 0 and variance 1.
So, we can say that our random variable $Z=\dfrac{X-\mu }{\sigma }$ follows standard normal distribution.
Now we need to find the value of $P\left( Z>0.8 \right)$. We can write it as,
$\Rightarrow P\left( Z>0.8 \right)=1-P\left( Z\le 0.8 \right)$
As $Z$ follows standard normal distribution, we can check the value of $P\left( Z\le 0.8 \right)$ from the normal distribution table, which is equal to 0.7881.
So, we get,
$\begin{align}
  & \Rightarrow P\left( Z>0.8 \right)=1-P\left( Z\le 0.8 \right) \\
 & \Rightarrow P\left( Z>0.8 \right)=1-0.7881 \\
 & \Rightarrow P\left( Z>0.8 \right)=0.2119 \\
\end{align}$
So, by substituting it in equation (2) we get that,
$\Rightarrow P\left( \dfrac{X-\mu }{\sigma }>0.8 \right)=0.2119$
So, by substituting it in equation (2) we get that,
$\begin{align}
  & \Rightarrow P\left( X>0.8\sigma +\mu \right)=0.2119 \\
 & \Rightarrow P\left( X>Jane's\ Score \right)=0.2119 \\
\end{align}$
So, as we need to find the number of students that score above Jane’s score let us multiply the probability with the total number of students.
$\begin{align}
  & \Rightarrow 500\times P\left( X>Jane's\ Score \right) \\
 & \Rightarrow 500\times 0.2119 \\
 & \therefore 105.95 \\
\end{align}$
So, we get the number of students getting a score above Jane’s score of 105.95. Rounding it off to the nearest integer we get 106.
So, the number of students who score above Jane is 106.

Hence the answer is Option B.

Note:
There is a possibility of one making a mistake while solving this problem by taking the value of probability, $P\left( Z>0.8 \right)$ from the table wrongly as 0.7881. But the table represents the value of $P\left( Z\le x \right)$ not $P\left( Z>x \right)$. So, we need to consider the value $P\left( Z>0.8 \right)$ as,
$\Rightarrow P\left( Z>0.8 \right)=1-P\left( Z\le 0.8 \right)$
Then we should look for the value of $P\left( Z\le 0.8 \right)$ from the normal distribution table.