Question

The average grade for an exam is 74, and the standard deviation is 7. The grades are curved to follow a normal distribution.

Answer 1

Answer:

0.9772 = 97.72% probability that the grade of a randomly selected students score is more than 60.

Step-by-step 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:

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

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 average grade for an exam is 74, and the standard deviation is 7.

This means that $\mu = 74, \sigma = 7$

What is the probability that the grade of a randomly selected students score is more than 60?

This is 1 subtracted by the pvalue of Z when X = 60. So

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

$Z = \frac{60 - 74}{7}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

1 - 0.0228 = 0.9772

0.9772 = 97.72% probability that the grade of a randomly selected students score is more than 60.