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Elina [12.6K]
3 years ago
6

A principal believes 50% of her music students reported their practice hours accurately. Based on that assumption, she ran 200 s

imulations 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
Mathematics
1 answer:
Alex17521 [72]3 years ago
4 0

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

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