Scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points. what is the minimum score you would need to be in the top 5%?

Using the normal distribution, it is found that the minimum score to be in the top 5% is of 86.4.

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In a normal distribution, the z-score is used.

is the mean and is the standard deviation.It measures how many standard deviations the measure is from the mean. Each z-score has an associated p-value, which is the percentile of the measure X.

In this problem:

Mean of 76.4, thus Standard deviation of 6.1, thus The top 5% is at least the 100 - 5 = 95th percentile, which is X when Z has a p-value of 0.95, so X when Z = 1.645.

The minimum score needed to be in the top 5% is of 86.4.

A similar problem is given at link

The top 5% of scores belong to the 95th percentile, which means the cutoff score is such that



Transforming to the standard normal distribution, you have



A left-tail probability of 95% corresponds to a z-score of about , which means the cutoff score must be around

Lol idk ask google. ejie