Question

Matthew participates in a study that is looking at how confident students at SUNY Albany are. The mean score on the scale is 50. The distribution has a standard deviation of 10 and is normally distributed. Matthew scores a 65. What percentage of people could be expected to score the same as Matthew or higher on this scale?
a) 93.32%
b) 6.68%
c) 0.07%
d) 43.32%

Answer 1

Answer:

b) 6.68%

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 mean score on the scale is 50. The distribution has a standard deviation of 10.

This means that $\mu = 50, \sigma = 10$

Matthew scores a 65. What percentage of people could be expected to score the same as Matthew or higher on this scale?

The proportion is 1 subtracted by the p-value of Z when X = 65. So

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

$Z = \frac{65 - 50}{10}$

$Z = 1.5$

$Z = 1.5$ has a p-value of 0.9332.

1 - 0.9332 = 0.0668

0.0668*100% = 6.68%

So the correct answer is given by option b.