Question

You recently took a statistics exam in a large class. The instructor tells the class that the scores were Normally distributed, with a mean of 72 (out of 100) and a standard deviation of 12. Your score was 96. You would like to know what proportion of students did better than you did. Unfortunately, your calculator batteries are dead and you must now rely on Normal tables. You need to find which of the following?
a. the area to the left of 2
b. the area to the right of 2
c. the area to the right of 96
d. the area to the left of 96

Answer 1

Answer:

b. the area to the right of 2

Step-by-step explanation:

Problems of normally distributed samples are 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, which is also the area to the left of Z. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X, which is the area to the right of Z.

In this problem:

$X = 96, \mu = 72, \sigma = 12$

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

$Z = \frac{96 - 72}{12}$

$Z = 2$

Percentage who did better:

P(Z > 2), which is the area to the right of 2.