Question

Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean $\mu =81$points and standard deviation$\sigma =15$ points.

a. Calculate the first- and third-quartile scores for this exam.

b. The middle $50\%$of the exam scores are between what two values?

Short answer

1. The first and third quartile scores for this exam are respectively $70.88and91.12$
2. The middle $50\%$of the exam scores are between$91.1173and70.8827$

Step-by-step solution

Given Information (Part a)

The total number of students who took exam is 2000.

The scores on the exam have an approximate normal distribution with a mean $\mu =81$points and standard deviation $\sigma =15$points.

Explanation (Part a)

Here,

$X~N\left(81,{15}^2\right)$

Let's find the first quartile score of the exam.

Here, the first quartile is equal to the ${25}^{th}$percentile and its represent the scores on the exam.

That means the probability $0.25$the scores on the exam and its less than$x$

localid="1649343919238" $P\left(x<Q_1\right)\mathit{=}\mathit{0}\mathit{.25}$

localid="1649343912971" $P\left[\frac{x-\mu }{6}<\frac{Q_1-\mu }{6}\right]\mathit{=}\mathit{0}\mathit{.25}$

localid="1649343907116" $\mathit{\Rightarrow }\frac{{\mathit{Q}}_{\mathit{1}}\mathit{-}\mathit{\mu }}{\mathit{6}}\mathit{=}\mathit{-}\mathit{0}\mathit{.6745}$

localid="1649343900499" $\mathit{\Rightarrow }\frac{{\mathit{Q}}_{\mathit{1}}\mathit{-}\mathit{81}}{\mathit{15}}\mathit{=}\mathit{-}\mathit{0}\mathit{.6745}$

localid="1649343891802" $Q_{\mathit{1}}\mathit{=}\mathit{15}x\mathit{-}\mathit{0}\mathit{.6745}\mathit{+}\mathit{81}$

localid="1649343881240" $Q_{\mathit{1}}\mathit{=}\mathit{70}\mathit{.8823}$

Let's find the third quartile score of the exam

$P\left(X<Q_3\right)\mathit{=}\mathit{0}\mathit{.}\mathit{75}$

$P\left[\frac{x-\mu }{6}<\frac{Q_3-81}{15}\right]\mathit{=}\mathit{0}\mathit{.}\mathit{75}$

$P\left(z<\frac{Q_3-81}{15}\right)\mathit{=}\mathit{0}\mathit{.}\mathit{75}$

$\mathit{\Rightarrow }\frac{{\mathit{Q}}_{\mathit{3}}\mathit{-}\mathit{81}}{\mathit{15}}\mathit{=}\mathit{0}\mathit{.}\mathit{6745}$

$Q_{\mathit{3}}\mathit{=}\mathit{15}\mathit{\times }\mathit{0}\mathit{.}\mathit{6745}\mathit{+}\mathit{81}$

$Q_{\mathit{3}}\mathit{=}\mathit{91}\mathit{.}\mathit{1175}$

Given Information (Part b)

The total number of students who took exam is 2000.

The scores on the exam have an approximate normal distribution with a mean $\mu =81$points and standard deviation$\sigma =15$ points.

Explanation (Part b)

The middle of $50\%$of the exam scores are lie between the third and first quartile.

That means

$0.75-0.25=0.50$