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

Answer:

$P(X>37)=P(\frac{X-\mu}{\sigma}>\frac{37-\mu}{\sigma})=P(Z>\frac{37-39.7}{2.3})=P(z>-1.173)$

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:

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

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

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

$P(X>37)=P(\frac{X-\mu}{\sigma}>\frac{37-\mu}{\sigma})=P(Z>\frac{37-39.7}{2.3})=P(z>-1.173)$

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:

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