Question

New York City $10$-km Run. As reported in Rumner's World magazine, the times of the finishers in the New York City $10$-km run are normally distributed with mean $61$ minutes and standard deviation $9$ minutes. Let $x$ denote finishing time for finishers in this race.

a. Sketch the distribution of the variable $x$.

b. Obtain the standardized version, $z$, of $x$.

c. Identify and sketch the distribution of $z$.

d. The percentage of finishers with times between $50$ and $70$ minutes is equal to the area under the standard normal curve between ---- and ---- .

e. The percentage of finishers with times less than $75$ minutes is equal to the area under the standard normal curve that lies to the ----- of ----- .

Short answer

a). The distribution of the variable $x$,

b). The standardized version of $z$,

$z=\frac{x-61}{9}$

c). The distribution of $z$will be

d). The percentage of finishers having time between $50$minutes and $70$minutes is equal to the area under the standard curve between the points $-1.22$and $1.00$.

e). The percentage of finishers having time less than $75$minutes is equal to the area under the standard curve between the points $0$and $1.56$.

Step-by-step solution

Part (a) Step 1: Given Information

Given data:

Mean $=61$minutes.

Standard deviation $=9$minutes

Part (a) Step 2: Explanation

The histogram of the distribution is roughly bell-shaped when the variable is roughly distributed regularly.

Part (b) Step 1: Given Information

Given data:

Mean $=61$ minutes.

Standard deviation $=9$ minutes.

Part (b) Step 2: Explanation

The histogram of the distribution is roughly bell-shaped when the variable is roughly distributed regularly.

The standardized version can be obtained:

$z=\frac{x-\text{mean}}{\text{standard deviation}}$

$z=\frac{x-61}{9}$

Part (c) Step 1: Given Information

Given data:

Mean $=61$ minutes.

Standard deviation $=9$ minutes.

Part (c) Step 2: Explanation

The distribution of $z$can be obtained,

Part (d) Step 1: Given Information

Given data:

Mean $=61$ minutes.

Standard deviation $=9$ minutes.

Part (d) Step 2: Explanation

The standardized version will be equal to

$z=\frac{x\text{-mean}}{\text{standard deviation}}$

$z=\frac{x-61}{9}$

For $x=50$,

$z=\frac{50-61}{9}$

$z=-1.22$

For $x=70$,

$z=\frac{70-61}{9}$

$z=1$

Part (e) Step 1: Given Information

Given data:

Mean $=61$ minutes.

Standard deviation $=9$ minutes.

Part (e) Step 2: Explanation

The standardized version can be

$z=\frac{x-\text{mean}}{\text{standard deviation}}$

$z=\frac{x-61}{9}$

When,$x=50$,

$z=\frac{75-61}{9}$

$z=1.56$