Question

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 _______%

Answer 1

Answer:

a) 50%

b) 5%

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.

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

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

$Z = \frac{60 - 60}{4}$

$Z = 0$

$Z = 0$ 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

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

$Z = \frac{66.5 - 60}{4}$

$Z = 1.63$

$Z = 1.63$ has a pvalue of 0.95

1 - 0.95 = 0.05

0.05*100% = 5%

5% probability.