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Ivan
3 years ago
11

Suppose the heights of professional horse jockeys are normally distributed with a mean of 62 in. and a standard deviation of 2 i

n.
Which group describes 16% of the population of horse jockeys?

Select each correct answer.




jockeys who are shorter than 58 in.

jockeys who are taller than 64 in.

jockeys who are shorter than 60 in.

jockeys who are between 60 in. and 64 in.
Mathematics
1 answer:
ch4aika [34]3 years ago
6 0
Mean = 62 in
SD = 2 in

At 16% = 0.16 and using Z tables,

Z≈ -0.99

x =(-0.99*2) + 62 ≈ 60 in

This represents jockeys who are shorter than 60 in.
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Answer:

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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.

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The correct answer is:

B.The score on Exam A is better, because the percentile for the Exam A score is higher.

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