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

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

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 = (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 = (X - \mu)/(\sigma)`

`Z = (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.