Question

The overhead reach distances of adult females are normally distributed with a mean of 195 cm and a standard deviation of 8 cm. a) Find the probability that an individual distance is greater than 192.8 cm.

Answer 1

Answer:

60.84%  probability that an individual distance is greater than 192.8 cm.

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 problem, we have that:

$\mu = 195, \sigma = 8$

Find the probability that an individual distance is greater than 192.8 cm.

This is 1 subtracted by the pvalue of Z when X = 192.8. So

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

$Z = \frac{192.8 - 195}{8}$

$Z = -0.275$

$Z = -0.275$ has a pvalue of 0.3916.

1 - 0.3916 = 0.6084

60.84%  probability that an individual distance is greater than 192.8 cm.