Answer:
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
and standard deviation
, the z-score of a measure X is given by:
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](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/ey4n9dzgcb9hfkti9yfo0h.png)
Proportion above 80:
1 subtracted by the p-value of Z when X = 80, so:
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/9mzqf4heaj8y4phlxu4lqj.png)
![Z = (80 - 67)/(8.1)](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/oeyr7y7tfim7malkylneux.png)
![Z = 1.6](https://img.qammunity.org/qa-images/2022/formulas/mathematics/college/pvu31kvmg1rnjcw02goe6e.png)
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%.