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

a. lie between \(73\) and \(80\)

b. are at least \(75\)

c. are at most \(90\)

The variable is normally distributed with mean \(68\) and standard deviation \(10\).

First find the probability of the variable \(X\) such that it is between \(73\) and \(80\) that is, \(P(73<X<80\)

Now calculating its value:

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

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

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

\(=P(0.5<Z<1.2)\)

\(=0.1935\)

The percentage will be \(0.1935\times100%=19.35%\)

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

First find the probability of the variable \(X\) such that it is more than \(75\) that is, \(P(75<X)\)

Now calculating its value:

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

\(=P(X-\mu>75-68)\)

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

\(=P(Z>0.7)\)

\(=0.999\)

The percentage will be \(0.999\times100%=99.9%\)

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

First find the probability of the variable \(X\) such that it is less than \(90\) that is, \(P(X<90)\)

Now calculating its value:

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

\(=P(X-\mu<90-68)\)

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

\(=P(Z<2.2)\)

\(=P(0<Z<2.2)\)

\(=0.0047\)

The percentage will be \(0.0047 \times 100%=0.47%\)

Hence, the percentage of all possible values of the variable that are at most \(90\) is \(0.47%\).