Chapter 2, Problem T2.12SPT

(a)

To determine

To calculate: Number of students who ran the mile in less than 6 minutes

Answer to Problem T2.12SPT

The number of students who ran the mile in less than 6 minutes is 802

Explanation of Solution

Given information:

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

Mean time of mile run of the male students = 7.11 minutes

Standard deviation of mile run time = 0.74 minutes

The times of mile run is normally distributed.

Formula used: $12,000\times \text{P (X < 6)}$

Calculation:

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

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

And standard deviation of mile run time be denoted by $\sigma$ .

So, $X~N\left(7.11,{{\displaystyle 0.74}}^2\right)$

P (X < 6) = $\text{P }\left(\frac{x-\mu }{\sigma }<\frac{6-7.11}{0.74}\right)$

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 6 minutes = 12000 $\times$ 0.06681

= 801.72 (802 approx.)

To determine

(b)

To calculate: Time taken by the slowest 10% of the students to run the mile

Answer to Problem T2.12SPT

Time taken by the slowest 10% of the students to run the mile is 8.1 minutes.

Explanation of Solution

Formula used:

P (X > T) = 0.10

Calculation:

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

Converting into standard normal

  $\begin{array}{l}\text{P }\left(\frac{x-\mu }{\sigma }>\frac{T-7.11}{0.74}\right)=0.10\\ \text{P }\left(z>\frac{T-7.11}{0.74}\right)=0.10\end{array}$

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

Upon comparing the values,

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

To determine

(c)

To calculate: Percentage of students who ran the mile between 400 and 500 seconds.

Answer to Problem T2.12SPT

Percentage of students who ran the mile between 400 and 500 seconds is 67.625%.

Explanation of Solution

Formula used: P (400 < Y < 500)

Calculation:

Converting seconds into minutes

  $P\left(\frac{400}{60}<X<\frac{500}{60}\right)$

P (6.667

  $\begin{array}{l}P\left(\frac{6.667-7.11}{0.74}<\frac{x-\mu }{\sigma }<\frac{8.333-7.11}{0.74}\right)\\ P(-0.599<z<1.653)\\ P(z<1.653)-P(z<-0.599)\\ P(z<1.653)-P(z>0.599)\\ P(z<1.653)-\left(1-P(z<0.599)\right)\\ \text{Using linear interpolation}\\ \text{P (z < 1}\text{.65) = 0}\text{.95053}\\ \text{P (z <1}\text{.66) = 0}\text{.95154}\\ \text{P (z < 1}\text{.653) = 0}\text{.95053 + (0}\text{.95154-0}\text{.95053) }\times \text{ 0}\text{.3 = 0}\text{.95083}\\ \text{P (z < 0}\text{.59) = 0}\text{.72240}\\ \text{P (z <0}\text{.60) = 0}\text{.0}\text{.72575}\\ \text{P (z < 0}\text{.599) = 0}\text{.72240 + (0}\text{.72575 - 0}\text{.72240) }\times 0.9=0.72542\end{array}$

Hence, the required probability is 0.95083 − (1 − 0.72542) = 0.67625 (= 67.625%)