Question

Suppose that x is normally distributed with a mean of 30 and a standard deviation of 3.What is P?

a) 0.469b) 0.956c) 0.493d) 0.490e) 0.466f) none of the above

Answer 1

If we assume that the deviation is
`\sigma=3` then the solution is:

`P(2.55<X<64.95)=P(-9.15<z<11.65)=P(z<11.65)-P(z<-9.15)`

f) None of the above

If we assume that the deviation is
`\sigma=15` then the solution is:

`P(2.55<X<64.95)=P(-1.83<z<2.33)=0.956`

b) 0.956

Explanation:

1) 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".

2) Solution to the problem

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

`X \sim N(30,3)`

Where
`\mu=30` and
`\sigma=3`

We are interested on this probability

`P(2.55<X<64.95)`

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(2.55<X<64.95)=P((2.55-\mu)/(\sigma)<(X-\mu)/(\sigma)<(64.95-\mu)/(\sigma))=P((2.55-30)/(3)<Z<(64.95-30)/(3))=P(-9.15<Z<11.65)`

And we can find this probability on this way:

`P(-9.15<z<11.65)=P(z<11.65)-P(z<-9.15)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-9.15<z<11.65)=0.99999`

If we assume that the deviation is
`\sigma=15` then the solution is:

`P(2.55<X<64.95)=P((2.55-\mu)/(\sigma)<(X-\mu)/(\sigma)<(64.95-\mu)/(\sigma))=P((2.55-30)/(15)<Z<(64.95-30)/(15))=P(-1.83<Z<2.33)`

And we can find this probability on this way:

`P(-1.83<z<2.33)=P(z<2.33)-P(z<-1.83)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-1.83<z<2.33)=0.956`