Question

A set of exam scores is normally distributed and has a mean of 74.4 and a standard deviation of 8.5. What is the probability that a randomly selected score will be between 69 and 77?

Answer = (round to four decimal places)

Note: Be careful...only use the Z Table here...do not use technology or the 68-95-99.7 Rule.

Answer 1

`P(69<x<77)=P(Z<0.306)-P(Z<-0.635)=0.6202-0.2627=0.3575`

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".

Let X the random variable who represent the exam scores for the population, and for this case we know the distribution for X is given by:

`X \sim N(74.4,8.5)`

Where
`\mu=74.4` and
`\sigma=8.5`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

What is the probability that a randomly selected score will be between 69 and 77?

For this case we can use the z score formula given by:

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

`P(69<x<77)=P((69-74.4)/(8.5)<Z<(77-74.4)/(8.5))=P(-0.635<Z<0.306)=P(Z<0.306)-P(Z<-0.635)=0.6202-0.2627=0.3575`

And that correspond with the 35.75% of the data.