Question

Let denote the time taken to run a road race. Suppose is approximately normally distributed with a mean of minutes and a standard deviation of minutes. If one runner is selected at random, what is the probability that this runner will complete this road race in to minutes? Round your answer to four decimal places. the tolerance is +/-2%

Answer 1

`P(221<X<237)=P((221-\mu)/(\sigma)<(X-\mu)/(\sigma)<(237-\mu)/(\sigma))=P((221-190)/(21)<Z<(237-190)/(21))=P(1.476<z<2.238)`

And we can find this probability with this difference:

`P(1.476<z<2.238)=P(z<2.238)-P(z<1.476)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(1.476<z<2.238)=P(z<2.238)-P(z<1.476)=0.987-0.924=0.063`

Explanation:

For this case we assume the following complete question: "Let x denote the time taken to run a road race. Suppose x is approximately normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete this road race in 221 to 237 minutes? "

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the time taken to run road race of a population, and for this case we know the distribution for X is given by:

`X \sim N(190,21)`

Where
`\mu=190` and
`\sigma=21`

We are interested on this probability

`P(221<X<237)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(221<X<237)=P((221-\mu)/(\sigma)<(X-\mu)/(\sigma)<(237-\mu)/(\sigma))=P((221-190)/(21)<Z<(237-190)/(21))=P(1.476<z<2.238)`

And we can find this probability with this difference:

`P(1.476<z<2.238)=P(z<2.238)-P(z<1.476)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(1.476<z<2.238)=P(z<2.238)-P(z<1.476)=0.987-0.924=0.063`