Question

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What percent of the students received a score below 64, if the mean grade on a test was 70 and the standard deviation was 6.​

Answer 1

Using the normal distribution, it is found that 15.87% of the students received a score below 64.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It 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, which is the percentile of X.

In this problem, the mean and the standard deviation are given by, respectively: $\mu = 70, \sigma = 6$.

The proportion of students who received a score below 64 is the p-value of Z when X = 64, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{64 - 70}{6}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

Hence, 15.87% of the students received a score below 64.

More can be learned about the normal distribution at https://brainly.com/question/24663213