Question

Suppose you are told that the grades of the nal exam of your class follows a normal distribution. The average score is 65 and the variance is 12. What is the the probability of a random classmate getting a score more than 80?

Answer 1

0% probability of a random classmate getting a score more than 80

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation(which is the square root of the variance)
`\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 question, we have that:

`\mu = 65, \sigma = √(12) = 3.4641`

What is the the probability of a random classmate getting a score more than 80?

This is 1 subtracted by the pvalue of Z when X = 80. So

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

`Z = (80 - 65)/(3.4641)`

`Z = 4.33`

`Z = 4.33` has a pvalue of 1

1 - 1 = 0

0% probability of a random classmate getting a score more than 80