Question

The distribution of grades in an introductory finance class is normally distributed, with an expected grade of 77. If the standard deviation of grades is 14, in what range would you expect 99.00 percent of the grades to fall? (Round answers to 2 decimal places, e.g. 15.25. Hint: Think in terms of what the expected highest and lowest scores would be for 99.00% of the students taking the exam.)

Answer 1

Answer:

99% of the scores fall between 40.95 and 113.05.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 77, \sigma = 14$.

99% of the students are between 99.5% and 0.05%. These values are

X when Z has a pvalue of 0.995.

So $Z = 2.575$.

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

$2.575 = \frac{X - 77}{14}$

$X - 77 = 2.575*14$

$X = 113.05$

X when Z has a pvalue of 0.005.

So $Z = -2.575$.

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

$-2.575 = \frac{X - 77}{14}$

$X - 77 = -2.575*14$

$X = 40.95$

99% of the scores fall between 40.95 and 113.05.