Question

Amy swims the 50-m backstroke. When in competition, her mean time is 39.7 seconds. And the standard deviation is 2.3 seconds. If she needs to swim faster than 37 seconds to earn a medal, what is the probability that she earns a medal? (Hint - use your normal distribution chart)

Answer 1

`P(X>37)=P((X-\mu)/(\sigma)>(37-\mu)/(\sigma))=P(Z>(37-39.7)/(2.3))=P(z>-1.173)`

And we can find this probability we can use this formula:

`P(z>-1.173)=1-P(z<-1.173)`

And using the normal standard distribution table or excel and we got:

`P(z>-1.173)=1-P(z<-1.173)=1-0.1204=0.8796`

Explanation:

Let X the random variable that represent the time so complete the 50 m of a population, and for this case we know the distribution for X is given by:

`X \sim N(39.7,2.3)`

Where
`\mu=39.7` and
`\sigma=2.3`

We are interested on this probability

`P(X>37)`

We can use the z score formula given by:

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

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

`P(X>37)=P((X-\mu)/(\sigma)>(37-\mu)/(\sigma))=P(Z>(37-39.7)/(2.3))=P(z>-1.173)`

And we can find this probability we can use this formula:

`P(z>-1.173)=1-P(z<-1.173)`

And using the normal standard distribution table or excel and we got:

`P(z>-1.173)=1-P(z<-1.173)=1-0.1204=0.8796`