A variable is normally distributed with a mean $6$and standard deviation $2$. Find the percentage of all possible values of the variable that

a. lie between $1$and $7$

b. exceed $5$

c. are less than $4$.

The given mean is $6$

Standard deviation is$2.$

The given data is written as

Mean, $\mu =6$

Standard deviation,$\sigma =2$

The random variable $X$has a $z-$score which is given as

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

Find the $z-$scores

For 1,

$z=\frac{1-6}{2}$

$=-2.5$

For $7$,

$z=\frac{7-6}{2}$

$=0.5$

Therefore the observations between $1$and $7$are the same as the $z-$scores between $-2.5$and $0.5$.

The difference between the numbers in Table II is then used to calculate the percentage of all observations:

$0.6915-0.0062=0.6853$

The given mean is $6$

Standard deviation is$2.$

Use the above formula and find the $z-$score

$z=\frac{5-6}{2}$

$=-0.5$

As a result, observations more than $5$correspond to z scores greater than $-0.5$.

The proportion of $z-$scores greater than $-0.5$is calculated in Table II of Appendix A

$1-0.3085=0.6915$

$=69.15\%.$

The given mean is $6$

Standard deviation is$2.$

The $z-$score is obtained by

$z=\frac{4-6}{2}$

$=-1.$

Therefore the observations less than $4$are the same $z-$scores less than $-1$. From Table II in Appendix A, the proportion of $z-$scores less than $-1$is

$0.1587=15.87\%.$