Question

Suppose the heights of women at a college are approximately Normally distributed with a mean of 66 inches and a population standard deviation of 1.5 inches. What height is at the 10 th ​percentile? Include an appropriately labeled sketch of the Normal curve to support your answer.

Answer 1

Answer:

$z=-1.28$

And if we solve for a we got

$a=66 -1.28*1.5=64.08$

So the value of height that separates the bottom 10% of data from the top 90% is 64.08.

The graph attached illustrate the situation.

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

$X \sim N(66,1.5)$  

Where $\mu=66$ and $\sigma=1.5$

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

$P(X>a)=0.90$   (a)

$P(X$   (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 0.10 of the area on the left and 0.90 of the area on the right it's z=-1.28. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05

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

$P(X$  

$P(z$

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

$z=-1.28$

And if we solve for a we got

$a=66 -1.28*1.5=64.08$

So the value of height that separates the bottom 10% of data from the top 90% is 64.08.

The graph attached illustrate the situation.