Question

The mean examination mark of a random sample of 1390 students is 67% with a standard deviation of 8.1%.

How many students scored above 80%? (Round to the nearest student)

Answer 1

76 students scored above 80%.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The mean examination mark of a random sample of 1390 students is 67% with a standard deviation of 8.1%.

This means that
`\mu = 67, \sigma = 8.1`

Proportion above 80:

1 subtracted by the p-value of Z when X = 80, so:

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

`Z = (80 - 67)/(8.1)`

`Z = 1.6`

`Z = 1.6` has a p-value of 0.9452.

1 - 0.9452 = 0.0548.

Out of 1390 students:

0.0548*1390 = 76

76 students scored above 80%.