Question

Suppose the heights of a population of people are normally distributed with a mean of 65.5 inches and a standard deviation of 2.6 inches. a. Find the probability that a randomly selected person from this population is between 64.2 and 67.2 inches tall. (5 points) b. What height separates the bottom 95% of data from the top 5%? (5 points)

Answer 1

Answer:

a) $P(64.2<X<67.2)=P(-0.50<z<0.65)=P(z<0.65)-P(-0.5)=0.742-0.309=0.434$

b) 69.764

Step-by-step explanation:

1) 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".

2) Part a

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:

$X \sim N(65.5,2.6)$

Where $\mu=65.5$ and $\sigma=2.6$

We are interested on this probability

$P(64.2<X<67.2)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

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

$P(64.2<X<67.2)=P(\frac{64.2-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{67.2-\mu}{\sigma})=P(\frac{64.2-65.5}{2.6}<Z<\frac{67.2-65.5}{2.6})=P(-0.50<z<0.65)$

And we can find this probability on this way:

$P(-0.50<z<0.65)=P(z<0.65)-P(-0.5)$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-0.50<z<0.65)=P(z<0.65)-P(-0.5)=0.742-0.309=0.434$

3) Part b

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:

$z=1.64<\frac{a-65.5}{2.6}$

And if we solve for a we got

$a=65.5 +1.64*2.6=69.764$

So the value of height that separates the bottom 95% of data from the top 5% is 69.764.