Question

A study of $12,000$ able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normally distributed with a mean of $7.11$ minutes and a standard deviation of $0.74$ minutes.$18$

a. About how many students ran the mile in less than $6$ minutes?

b. Approximately how long did it take the slowest $10\%$ of students to run the mile?

c. Suppose that these mile run times are converted from minutes to seconds. Estimate the percent of students who ran the mile in between $400$ and $500$ seconds.

Short answer

Part (a) The number of students who ran the mile in less than $6$minutes is $802$

Part (b) Time taken by the slowest $10\%$of the students to run the mile is $8.1$minutes.

Part (c) Percentage of students who ran the mile between $400$ and $500$ seconds is $67.625\%.$

Step-by-step solution

Part (a) Step 1: Given information

Total number of male students in the study$=12,000$

Male students' mile run average time $=7.11minutes$

The mile run time's standard deviation $=0.74minutes$

Mile run timings are generally spread.

Part (a) Step 2: Concept

$12,000\times P(X<6)$

Part (a) Step 3: Calculation

Let $X$be the University of Illinois mile run time.

Let the mean mile run time be denoted by $\mu$

And the standard deviation of mile run time is denoted by $\sigma$

So,

$$\begin{gathered} X~N(7.11,{0.74}^2) \\ P(X<6)=P(\frac{x-\mu }{\sigma }<\frac{6-7.11}{0.74}) \end{gathered}$$

$P(z<-1.5)=P(z>1.5)$as per symmetry

So, $P(z>1.5)=1-P(z<1.5)$

From $z$tables, $P(z<1.5)=0.93319$

Therefore, $P(z>1.5)=1-0.93319=0.06681$

The number of students who took less than

$$\begin{gathered} 6minutes=12000\times 0.06681 \\ =801.72(802approx.) \end{gathered}$$

Part (b) Step 2: Concept

$P(X>T)=0.10$

Part (b) Step 2: Calculatrion

Let $T$ be the time above which there are slowest $10\%$ of the students.

$$\begin{gathered} P(\frac{x-\mu }{\sigma }>\frac{T-7.11}{0.74})=0.10 \\ P(z>\frac{T-7.11}{0.74})=0.10 \end{gathered}$$

Converting into standard normal

From $z$ tables, $P(z>1.2816)=0.10$

Upon comparing the values,

$$\begin{gathered} 1.2816=\frac{T-7.11}{0.74} \\ T=7.11+0.74\times 1.2816=8.058 \end{gathered}$$

Part (c) Step 2: Concept

P (400 < Y < 500)

Part (c) Step 2: Calculation

Converting seconds into minutes

$$\begin{gathered} P(\frac{400}{60}<X<\frac{500}{60}) \\ P(\frac{6.667-7.11}{0.74}<\frac{x-\mu }{\sigma }<\frac{8.333-7.11}{0.74}) \\ P(-0.599<z<1.653) \\ P(z<1.653)-P(z<-0.599) \\ P(z<1.653)-P(z>0.599) \\ P(z<1.653)-(1-P(z<0.599\left)\right) \end{gathered}$$

Using linear interpolation

$$\begin{gathered} P(z<1.65)=0.95053 \\ P(z<1.66)=0.95154 \\ P(z<1.653)=0.95053+(0.95154-0.95053)\times 0.3=0.95083 \\ P(z<0.59)=0.72240 \\ P(z<060)=0072575 \end{gathered}$$

$P(z<0.599)=0.72240+(0.72575-0.72240)\times 0.9=0.72542$

Hence, the required probability is $0.95083-(1-0.72542)=0.67625(=67.625\%)$