Question

Suppose that the duration of a routine doctor's visit is known to be normally distributed with a mean of 21 minutes and a standard deviation of seven minutes. If one of the visits is randomly chosen, what is the probability that it lasted at least 24 minute?

Answer 1

Answer:

33.36% probability that it lasted at least 24 minutes

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 21, \sigma = 7$

If one of the visits is randomly chosen, what is the probability that it lasted at least 24 minute?

This is 1 subtracted by the pvalue of Z when X = 24. So

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

$Z = \frac{24 - 21}{7}$

$Z = 0.43$

$Z = 0.43$ has a pvalue of 0.6664

1 - 0.6664 = 0.3336

33.36% probability that it lasted at least 24 minutes