Question

Last year, the personal best high jumps of
track athletes in a nearby state were
normally distributed with a mean of
208 cm and a standard deviation of
19 cm What is the probability that a
randomly selected high jumper has a
personal best between 227 and 265 cm?

Answer 1

Using the normal distribution, it is found that there is a 0.84 = 84% probability that a randomly selected high jumper has a personal best between 227 and 265 cm.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 208, \sigma = 19$

The probability that a randomly selected high jumper has a personal best between 227 and 265 cm is the p-value of Z when X = 265 subtracted by the p-value of Z when X = 227, hence:

X = 265:

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

$Z = \frac{265 - 208}{19}$

Z = 3

Z = 3 has a p-value of 0.9987.

X = 227:

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

$Z = \frac{227 - 208}{19}$

Z = 1

Z = 1 has a p-value of 0.1587.

0.9987 - 0.1587 = 0.84.

0.84 = 84% probability that a randomly selected high jumper has a personal best between 227 and 265 cm.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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