Question

64. Suppose that a category of world-class runners is known to run a marathon ( 26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let $\stackrel{-}{X}$be the average of the 49 races.

a. role="math" localid="1652281768411" $\stackrel{-}{X}~$(___)

b. Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons.

c. Find the 80th percentile for the average of these 49 marathons.

d. Find the median of the average running times.

Short answer

a. $\overline{X}~N(145,2)$

b.$0.624$

c.$146.68$

d.$145$minutes

Step-by-step solution

Given Information

At the point when we're uncertain about the result of an occasion, we can discuss the probabilities of specific results and how likely they are to occur. The number zero indicates the uncertainty of the event and one indicates the certainity.

Explanation Part (a)

Given,

The standard deviation $\sigma$is $14$minutes.

Sample size n = $49$

Average time (mean)${\mu }_x=145$minutes

The average of $49$races is $\stackrel{-}{X}$

Using the formula and substituting the values,

$\overline{X}~N\left({\mu }_x,\frac{{\sigma }_x}{\sqrt{n}}\right)$

$\overline{X}~N\left(145,\frac{14}{\sqrt{49}}\right)$

Hencerole="math" localid="1652282570889" $\stackrel{-}{X}~N(145,2)$

Explanation Part (b)

We know,

$\stackrel{-}{X}~N(145,2)$

where$\stackrel{-}{X}$is the average of$49$races

Finding the probability that the runner will average between $142$and$146$minutes in these $49$marathons is,

Using the calculator,

role="math" localid="1652282899504" $$\begin{gathered} P(142<\overline{x}<146)=\text{normalcdf}(142,146,145,2) \\ =0.624 \end{gathered}$$

Explanation Part (c)

We know,

$\stackrel{-}{X}~N(145,2)$

where $\stackrel{-}{X}$is the average of $49$races.

For the average of these$49$marathons the $80th$percentile is,

$P(\overline{x}<k)=0.80$

Using the calculator to find k,

$\text{invnorm}(0.80,145,2)=146.68$

Hence the $80th$percentile is$146.68$

Explanation Part (d)

Given,

$\stackrel{-}{X}~N(145,2)$

where $\stackrel{-}{X}$is the average of $49$races

We know the median is = the mean of the random variable in a normal distribution

Hence, median = mean = $145$

The median of the average running times is$145$min