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J.V. Lin is an Olympic athlete competing in the women's javelin throw. The distances of her successful throws (in meters) are normally distributed with mean distance 60 meters and standard deviation 4 meters. (Round to the nearest percent and enter each of your answers as a whole number between 0 and 100.) (a) (2 points) What is the probability that on her first throw, J.v. Lin beats her mean distance? probability = ________%(b) (3 points) The current leader in the Olympics finals has thrown a distance of 66.5 meters. J.V. Lin has one attempt left to beat the leader's throw. What's the probability that Lin beats this throw? probability _______%

User Kiku
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1 Answer

9 votes
9 votes

Answer:

a) 50%

b) 5%

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 = (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.

Normally distributed with mean distance 60 meters and standard deviation 4 meters.

This means that
\mu = 60, \sigma = 4

(a) (2 points) What is the probability that on her first throw, J.v. Lin beats her mean distance?

This is, as a proportion, 1 subtracted by the pvalue of Z when X = 60. So


Z = (X - \mu)/(\sigma)


Z = (60 - 60)/(4)


Z = 0


Z = 0 has a pvalue of 0.5

1 - 0.5 = 0.5

0.5*100% = 50%

50% probability.

(b) (3 points) The current leader in the Olympics finals has thrown a distance of 66.5 meters. J.V. Lin has one attempt left to beat the leader's throw. What's the probability that Lin beats this throw?

This is 1 subtracted by the pvalue of Z when X = 66.5. So


Z = (X - \mu)/(\sigma)


Z = (66.5 - 60)/(4)


Z = 1.63


Z = 1.63 has a pvalue of 0.95

1 - 0.95 = 0.05

0.05*100% = 5%

5% probability.

User Nazaria
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