A race director is preparing for an upcoming marathon and estimates that the mean time to
finish is 310 minutes. Assume
that the times are normally distributed, with a standard deviation of 50 minutes.
Use a standard normal table or a calculator to find the percentage of times that are longer than 236 minutes. For your
intermediate computations, use four or more decimal places. Give your final answer to two decimal places (for example
98.23%).

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raphealnwobi raphealnwobi · Answer 1 · 2021-12-15T10:08:39+01:00

93.06% of the race is longer than 236 minutes.

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]

Given that μ = 310, σ = 50

For x > 236:

[tex]z=\frac{236-310}{50}=-1.48[/tex]

From the normal distribution table:

P(x > 236) = P(z > -1.48) = 1 - P(z < -1.48) = 1 - 0.0694 = 0.9306 = 93.06%

93.06% of the race is longer than 236 minutes.

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