Question

When Elena throws a shot put during track and field practice, it travels an average of 26 feet with a standard deviation of 3 feet, and the distribution of lengths is approximately normal. Approximately what percentage of her shots will travel less than 23 feet? *

Answer 1

Answer:

15.87%, that is, approxiately 16% of of her shots will travel less than 23 feet.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 26, \sigma = 3$

Approximately what percentage of her shots will travel less than 23 feet?

This is the pvalue of Z when X = 23. So

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

$Z = \frac{23 - 26}{3}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

15.87%, that is, approxiately 16% of of her shots will travel less than 23 feet.