Q. 6.91 - Elementary Statistics

Question

A variable is normally distributed with a mean $10$ and standard deviation$3$. Find the percentage of all possible values of the variable that

a. lie between $6$ and $7$.

b. are at least $10$ .

c. are at most $17.5$ .

Step-by-Step Solution

Answer

a) The percentage of all possible values of the variable that lie between $6$ and $7$ is  $0.06\%.$

b) The percentage of all possible values of the variable that are at least $10$ is $99.96\%$

c) The percentage of all possible values of the variable that are at most  $17.5$ is $0.62\%.$

Step 1: Given Information (Part a)

The variable is normally distributed with a mean $10$ and a standard deviation $3$.

$\mu =10;\sigma =3$

Step 2: Explanation (Part a)

First, find the probability of the variable $X$ such that it is between $6$ and $7$ that is, $P(6<X<7)$

Now calculating its value:

$P(6<X<7)=P(6-\mu <X-\mu <7-\mu )$

                      $=P(6-10X-\mu <7-10)$

                      $=P\left(\frac{6-10}{3}<\frac{X-\mu }{\sigma }<\frac{7-10}{3}\right)$

                      $=P(-1.33<Z<-1)$$=P(-1.33<Z<-1)$

                      $=P(1<Z<1.33)$

                      $=0.0006$

The percentage will be $0.0006\times 100\%=0.06\%$.

Step 1: Given Information (Part b)

The variable is normally distributed with a mean $10$ and a standard deviation$3 .$

$\mu =10;\sigma =3$

Step 2: Explanation (Part b)

First, find the probability of the variable $X$ such that it is more than  $10$ that is, $P(10<X).$

Now calculating its value:

$P(X>10)=P(X-\mu >10-\mu )$

$=P(X-\mu >10-10)$

$=P\left(\frac{X-\mu }{\sigma }>\frac{10-10}{3}\right)$

$=P(Z>0)$

$=0.9996$

The percentage will be $0.9996\times 100\%=99.96\%$ .

Step 1: Given Information (Part c)

The variable is normally distributed with a mean of $10$ and a standard deviation of $3$.

$\mu =10;\sigma =3$

Step 2: Explanation (Part c)

First, find the probability of the variable $X$ such that it is less than  $17.5$ that is, $P(X<17.5)$

Now calculating its value:

$P(X<17.5)=P(X-\mu <17.5-\mu )$

$=P(X-\mu <17.5-10)$

$=P\left(\frac{X-\mu }{\sigma }<\frac{17.5-10}{3}\right)$

$=P(Z<2.5)$

$=P(0<Z<2.5)$

$=0.0062$

The percentage will be  $0.0062\times 100\%=0.62\%$.