Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 79 and a standard deviation of 6. The top 15% of all scores have been designated As. Your score is 89. Did you earn an A

Respuesta :

Answer:

You earned an A.

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. 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:

[tex]\mu = 79, \sigma = 6[/tex]

The top 15% of all scores have been designated As.

This means that if Z for the score has a pvalue of 1-0.15 = 0.85 or higher, the score is designated as A.

Your score is 89. Did you earn an A?

We have to find the pvalue of Z when X = 89. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{89 - 79}{6}[/tex]

[tex]Z = 1.67[/tex]

[tex]Z = 1.67[/tex] has a pvalue of 0.9525. So yes, you earned an A.