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?

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 and standard deviation , the zscore of a measure X is given by:

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:

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

has a pvalue of 0.6664

1 - 0.6664 = 0.3336

33.36% probability that it lasted at least 24 minutes

ian gone lie this got me confused

step-by-step explanation:

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.