Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a standard deviation of 1900 hours and a mean life span of 19,000 hours. If a monitor is selected at random, find the probability that the life span of the monitor will be more than 15,579 hours. Round your answer to four decimal places.

0.9641 = 96.41% probability that the life span of the monitor will be more than 15,579 hours.

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:

Find the probability that the life span of the monitor will be more than 15,579 hours.

This is 1 subtracted by the pvalue of Z when X = 15579. So

has a pvalue of 0.0359

1 - 0.0359 = 0.9641

0.9641 = 96.41% probability that the life span of the monitor will be more than 15,579 hours.

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step-by-step explanation: