Question

Assume the random variable X is normally distributed with mean

mu equals 50μ=50

and standard deviation

sigma equals 7σ=7.

Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded.

Upper P left parenthesis 34 less than Upper X less than 63 right parenthesis

P(34

Answer 1

`P(34<X<63)=P((34-\mu)/(\sigma)<(X-\mu)/(\sigma)<(63-\mu)/(\sigma))=P((34-50)/(7)<Z<(63-50)/(7))=P(-2.286<z<1.857)`

And we can find this probability with this difference and using the normal standard table or excel:

`P(-2.286<z<1.857)=P(z<1.857)-P(z<-2.286)=0.968-0.0111= 0.9569`

And the result is illustrated in the figure attached.

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

`X \sim N(50,7)`

Where
`\mu=50` and
`\sigma=7`

We are interested on this probability

`P(34<X<63)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(34<X<63)=P((34-\mu)/(\sigma)<(X-\mu)/(\sigma)<(63-\mu)/(\sigma))=P((34-50)/(7)<Z<(63-50)/(7))=P(-2.286<z<1.857)`

And we can find this probability with this difference and using the normal standard table or excel:

`P(-2.286<z<1.857)=P(z<1.857)-P(z<-2.286)=0.968-0.0111= 0.9569`

And the result is illustrated in the figure attached.