Question

A population of values has a normal distribution with μ = 157.7 and σ = 57 . You intend to draw a random sample of size n = 24 . Find P95, which is the mean separating the bottom 95% means from the top 5% means.

Answer 1

Answer:

$z=1.64<\frac{a-157.7}{57}$

And if we solve for a we got

$a=157.7 +1.64*57=251.18$

So the value of height that separates the bottom 95% of data from the top 5% is 251.18.  

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

$X \sim N(157.7,57)$  

Where $\mu=157.7$ and $\sigma=57$

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

$P(X>a)=0.05$   (a)

$P(X<a)=0.95$   (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 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. 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:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.95$  

$P(z<\frac{a-\mu}{\sigma})=0.95$

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

$z=1.64<\frac{a-157.7}{57}$

And if we solve for a we got

$a=157.7 +1.64*57=251.18$

So the value of height that separates the bottom 95% of data from the top 5% is 251.18.