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 X-be the average of the 49 races.

a. role="math" localid="1652281768411" 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

Expert verified

a. X¯~N(145,2)

b.0.624

c.146.68

d.145minutes

Step by step solution

01

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.

02

Explanation Part (a)

Given,

The standard deviation σis 14minutes.

Sample size n = 49

Average time (mean)μx=145minutes

The average of 49races is X-

Using the formula and substituting the values,

X¯~Nμx,σxn

X¯~N145,1449

Hencerole="math" localid="1652282570889" X-~N(145,2)

03

Explanation Part (b)

We know,

X-~N(145,2)

whereX-is the average of49races

Finding the probability that the runner will average between 142and146minutes in these 49marathons is,

Using the calculator,

role="math" localid="1652282899504" P(142<x¯<146)=normalcdf(142,146,145,2)=0.624

04

Explanation Part (c)

We know,

X-~N(145,2)

where X-is the average of 49races.

For the average of these49marathons the 80thpercentile is,

P(x¯<k)=0.80

Using the calculator to find k,

invnorm(0.80,145,2)=146.68

Hence the 80thpercentile is146.68

05

Explanation Part (d)

Given,

X-~N(145,2)

where X-is the average of 49races

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 is145min

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