Suppose that X is a normal random variable with mean 5. If P(X>9)=0.2, approximately what is Var(X)?

This question aims to find the probability of a normally distributed random variable $X$. A random variable is one whose value is determined by the results of a statistical experiment.

The normal distribution, also known as the Gaussian distribution or the z-distribution, has a mean of zero and a standard deviation of one. Data in a normal distribution is symmetrically distributed and has no skew. The data takes the shape of a bell when plotted on a graph, with most values grouping around a central region and scattering off as they move away from the center.

The two characteristics such as mean and standard deviation define the graph of the normal distribution. The mean/average is the maximum of the graph, whereas the standard deviation measures the amount of spread away from the mean.

Expert Answer

Let $\mu$ and $\sigma$ be the mean and standard deviation of the random variable $X$. According to the question:

$\mu =5$, $P(X>9)=0.2$  and  we have to find  Var(X) $={\sigma }^2$.

Since, $P(X>9)=0.2$

$⟹P(X<9)=1-0.2=0.8$

$⟹P(Z<{\displaystyle \frac{x-\mu }{\sigma }})=0.8$

$⟹P(Z<{\displaystyle \frac{9-5}{\sigma }})=0.8$

$⟹\varphi \left({\displaystyle \frac{9-5}{\sigma }}\right)=0.8$

So, by inverse use of the $z-$ table, when $\varphi (z)=0.8$ then $z\approx 0.84$. And hence:

${\displaystyle \frac{9-5}{\sigma }}=0.84$

${\displaystyle \frac{4}{\sigma }}=0.84$

$\sigma ={\displaystyle \frac{4}{0.84}}=4.76$

Therefore,  Var (X) $={\sigma }^2=(4.76{)}^2=22.66$

Example 1

Consider $X$ as a normally  distributed random variable with $\mu =22$ and $\sigma =3$. Find $P(X<23)$, $P(X>19)$  and  $P(25<X<30)$.

Solution

Here, $\mu =22$ and $\sigma =3$

Therefore,  $P(X<23)=P(Z<{\displaystyle \frac{X-\mu }{\sigma }})$

$⟹P(Z<{\displaystyle \frac{23-22}{3}})=P(Z<0.33)=0.6293$

Now, $P(X>19)=P(Z>{\displaystyle \frac{X-\mu }{\sigma }})$

$⟹P(Z>{\displaystyle \frac{19-22}{3}})=P(Z>-1)$

$P(Z>-1)=1-P(Z<-1)=1-0.1587=0.8413$

Also,  $P(25<X<30)=P({\displaystyle \frac{25-22}{3}}<Z<{\displaystyle \frac{30-22}{3}})$

$⟹P(1<Z<2.66)=P(0<Z<2.66)-P(1<Z<0)=0.9961-0.8413=0.1548$

Area under the normal curve between $25$ and $30$

Example 2

The time between battery charges for some specific types of computers is normally distributed, with a mean of $30$ hours and a standard deviation of $12$ hours.  Alice has one of these computer systems and is curious about the likelihood that the time will be between $60$ and $80$ hours.

Solution

Here, $\mu =30$  and $\sigma =12$

To find: $P(60<X<80)$

Now, $P(60<X<80)=P({\displaystyle \frac{60-30}{12}}<Z<{\displaystyle \frac{80-30}{12}})$

$⟹P(2.5<Z<4.16)=P(0<Z<4.16)-P(2.5<Z<0)$

$=0.4998-0.4938=0.0060$

Example 3

A normal distribution model with a mean of $6$ cm and a standard deviation of $0.03$ cm is used to approximate the length of similar components produced by a company. If one component is selected randomly, what is the likelihood that this component’s length is between $5.89$ and $6.03$ cm?

Solution

Given,  $\mu =6$ and $\sigma =0.03$

To find: $P(5.89<X<6.03)$

Now, $P(5.89<X<6.03)=P({\displaystyle \frac{5.89-6}{0.03}}<Z<{\displaystyle \frac{6.03-6}{0.03}})$

$⟹P(-3.66<Z<1)=P(-3.66<Z<0)+P(0<Z<1)$

$=0.0002+0.8413=0.8415$

                   Images/mathematical drawings are created with GeoGebra.

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