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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]X = 96, \mu = 72, \sigma = 12[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{96 - 72}{12}[/tex]
[tex]Z = 2[/tex]
Percentage who did better:
P(Z > 2), which is the area to the right of 2.
