Question

In a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of 69.5inches and a standard deviation of 3.0inches. A study participant is randomly selected. Complete parts​ (a) through​ (d) below.

a. Find the probability that a study participant has a height that is less than 66 inches.
b. Find the probability that a study participant has a height that is between 66 and 71 inches.
c. Find the probability that a study participant has a height that is more than 71 inches.
d. Identify any unusual events. Explain your reasoning.

Answer 1

Answer:

a) $P(X<66)=P(\frac{X-\mu}{\sigma}<\frac{66-\mu}{\sigma})=P(Z<\frac{66-69.5}{3.0})=P(z<-1.167)$

And we can find this probability using the normal standard table:

$P(z<-1.167)=0.122$

b) $P(66<X<71)=P(\frac{66-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{71-\mu}{\sigma})=P(\frac{66-69.5}{3}<Z<\frac{71-69.5}{3})=P(-1.167<z<0.5)$

And we can find this probability with thi difference:

$P(-1.167<z<0.5)=P(z<0.5)-P(z<-1.167)$

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

$P(-1.167<z<0.5)=P(z<0.5)-P(z<-1.167)=0.691-0.122=0.569$

c) $P(X>71)=P(\frac{X-\mu}{\sigma}>\frac{71-\mu}{\sigma})=P(Z>\frac{71-69.5}{3.0})=P(z>0.5)$

And we can find this probability using the complement rule and the normal standard table:

$P(z>0.5)=1-P(z<0.5) =1-0.691=0.309$

d) $\mu - 2*\sigma = 69.5 -2*3=63.5$

$\mu + 2*\sigma = 69.5 + 2*3=75.5$

If a value is lower than 63.5 or higher than 75.5 we can consider it as unusual

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".  

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:

$X \sim N(69.5,3.0)$  

Where $\mu=69.5$ and $\sigma=3.0$

We are interested on this probability

$P(X<66)$

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:

$P(X<66)=P(\frac{X-\mu}{\sigma}<\frac{66-\mu}{\sigma})=P(Z<\frac{66-69.5}{3.0})=P(z<-1.167)$

And we can find this probability using the normal standard table:

$P(z<-1.167)=0.122$

Part b

$P(66<X<71)=P(\frac{66-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{71-\mu}{\sigma})=P(\frac{66-69.5}{3}<Z<\frac{71-69.5}{3})=P(-1.167<z<0.5)$

And we can find this probability with thi difference:

$P(-1.167<z<0.5)=P(z<0.5)-P(z<-1.167)$

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

$P(-1.167<z<0.5)=P(z<0.5)-P(z<-1.167)=0.691-0.122=0.569$

Part c

$P(X>71)=P(\frac{X-\mu}{\sigma}>\frac{71-\mu}{\sigma})=P(Z>\frac{71-69.5}{3.0})=P(z>0.5)$

And we can find this probability using the complement rule and the normal standard table:

$P(z>0.5)=1-P(z<0.5) =1-0.691=0.309$

Part d

For this case we can consider as unusual events values above/ below two deviations within the mean:

$\mu - 2*\sigma = 69.5 -2*3=63.5$

$\mu + 2*\sigma = 69.5 + 2*3=75.5$

If a value is lower than 63.5 or higher than 75.5 we can consider it as unusual