Question

Suppose the scores on an exam are normally distributed with a mean μ = 75 points, and standard deviation σ = 8 points.

The instructor wanted to "pass" anyone who scored above 69. What proportion of exams will have passing scores?

Answer 1

`P(X>69)=P((X-\mu)/(\sigma)>(69-\mu)/(\sigma))=P(Z>(69-75)/(8))=P(Z>-0.75)`

And we can find this probability on this way:

`P(Z>-0.75)=1-P(Z<-0.75)= 1-0.227= 0.773`

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 problm

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

`X \sim N(75,8)`

Where
`\mu=75` and
`\sigma=8`

We are interested on this probability

`P(X>69)`

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(X>69)=P((X-\mu)/(\sigma)>(69-\mu)/(\sigma))=P(Z>(69-75)/(8))=P(Z>-0.75)`

And we can find this probability on this way:

`P(Z>-0.75)=1-P(Z<-0.75)= 1-0.227= 0.773`