Question

A principal believes 50% of her music students reported their practice hours accurately. Based on that assumption, she ran 200 simulations of random samples. She found the mean of each run. Then she found the mean, 48%, and standard deviation of all the means, 18%, to create a normal distribution.
According to that distribution, which option below gives the best description of how likely is it that a single survey would return a mean of 30%?

A. extremely likely
B. likely
C. unlikely
D. extremely unlikely

Answer 1

Answer:

C. unlikely

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.

A probability is said to be extremely likely if it is 95% or higher, and extremely unlikely if it is 5% or lower. A probabilty higher than 50% and lower than 95% is said to be likely, and higher than 5% and lower than 50% is said to be unlikely.

In this problem, we have that:

$\mu = 0.48, \sigma = 0.18$

How likely is it that a single survey would return a mean of 30%?

We have to find the pvalue of Z when X = 0.30.

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

$Z = \frac{0.30 - 0.48}{0.18}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

So the correct answer is:

C. unlikely