Question

A physical fitness association is including the mile run in its secondary-school fitness test. The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 470 seconds and a standard deviation of 60 seconds. The fitness association wants to recognize the boys whose times are among the top (or fastest) 10% with certificates of recognition. What time would the boys need to beat in order to earn a certificate of recognition from the fitness association

Answer 1

Answer:

$z=-1.28<\frac{a-470}{60}$

And if we solve for a we got

$a=470 -1.28*60=393.2$

So the value of height that separates the bottom 10% (Fastest 10%) of data from the top 90% is 393.2.

so if the time is 393.2 or lower a boy would earn a certificate of recognition from the fitness association

Step-by-step explanation:

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 for this event of a population, and for this case we know the distribution for X is given by:

$X \sim N(470,60)$  

Where $\mu=470$ and $\sigma=60$

NOTE: For this case the fastest 10% is on the lower tail of the distribution

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.9$   (a)

$P(X<a)=0.1$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.1$  

$P(z<\frac{a-\mu}{\sigma})=0.1$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-1.28<\frac{a-470}{60}$

And if we solve for a we got

$a=470 -1.28*60=393.2$

So the value of height that separates the bottom 10% (Fastest 10%) of data from the top 90% is 393.2.

so if the time is 393.2 or lower a boy would  earn a certificate of recognition from the fitness association