Question

A statistics teacher believes that the final exam grades for her class have a normal distribution with a mean of 82 and a standard deviation of 8. Answer the following: if x represents a possible test score for this population, find P(x >92)

Answer 1

`P(X>92)=P((X-\mu)/(\sigma)>(92-\mu)/(\sigma))=P(Z>(92-82)/(8))=P(z>1.25)`

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

`P(z>1.25)=1-P(z<1.25)=1-0.894=0.106`

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 final exam grades of a population, and for this case we know the distribution for X is given by:

`X \sim N(82,8)`

Where
`\mu=82` and
`\sigma=8`

We are interested on this probability

`P(X>92)`

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>92)=P((X-\mu)/(\sigma)>(92-\mu)/(\sigma))=P(Z>(92-82)/(8))=P(z>1.25)`

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

`P(z>1.25)=1-P(z<1.25)=1-0.894=0.106`

Answer 2

P(x >92) = 0.1056

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 82, \sigma 8`

P(x >92)

This is 1 subtracted by the pvalue of Z when X = 92. So

`Z = (X - \mu)/(\sigma)`

`Z = (92 - 82)/(8)`

`Z = 1.25`

`Z = 1.25` has a pvalue of 0.8944

1 - 0.8944 = 0.1056

P(x >92) = 0.1056