Q. 6.90 - Elementary Statistics

Question

A variable is normally distributed with a mean $68$ and standard deviation $10$. Find the percentage of all possible values of the variable that

a. lie between $73$ and $80$.

h. are at least $75$.

c. are at most $90$.

Step-by-Step Solution

Answer

(a)  The percentage of all possible values of the variable that lie between $73$ and $80$ is $19.35\%.$

(b) The percentage of all possible values of the variable that are at least $75$ is $99.9\%.$

(c) The percentage of all possible values of the variable that are at most $90$ is $0.47\%.$

Part (a) Step 1: Given information

The given mean is $68$

Standard deviation is $10.$

Part (a) Step 2: Explanation

The given data is 

Mean, $\mu =68$

Standard deviation, $\sigma =10$

If X is a continuous random variable with mean $\mu$ and standard deviation $\sigma$, then a normal curve with X has the following equation:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0.$

Moreover, the equation of a normal curve with random variable $Z$ is as follows:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

Using the values, Find $P(73<X<80)$

$P(73<X<80)=P(73-\mu <X-\mu <80-\mu )$

                          $=P(73-68<X-\mu <80-68)$

                          $=P\left(\frac{73-68}{10}<\frac{X-\mu }{\sigma }<\frac{80-68}{10}\right)$   

                          $=P(0.5<Z<1.2)$

                           $=0.1935$

                           $=19.35\%.$

Part (b) Step 1: Given information

The given mean is $68$

Standard deviation is $10.$

Part (b) Step 2: Explanation

If X is a continuous random variable with mean $\mu$ and standard deviation $\sigma$, then a normal curve with X has the following equation:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0.$

Moreover, the equation of a normal curve with random variable $Z$ is as follows:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

then find, 

$P(X>75)=P(X-\mu >75-\mu )$

                  $=P(X-\mu >75-68)$

                  $=P\left(\frac{X-\mu }{\sigma }>\frac{75-68}{10}\right)$

                   $=P(Z>0.7)$

                   $=0.999$

                   $=99.9\%.$

Part (c) Step 1: Given information

The given mean is $68$

Standard deviation is $10$

Part (c) Step 2: Explanation

If X is a continuous random variable with mean $\mu$ and standard deviation $\sigma$, then a normal curve with X has the following equation:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0.$

Moreover, the equation of a normal curve with random variable $Z$ is as follows:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

Then find, 

$P(X<90)=P(X-\mu <90-\mu )$

                  $=P(X-\mu <90-68)$

                 $=P\left(\frac{X-\mu }{\sigma }<\frac{90-68}{10}\right)$

                 $=P(Z<2.2)$

                 $=P(0<Z<2.2)$

                 $=0.0047$

                 $=0.47\%.$