Battery lifetime is normally distributed for large samples. The mean lifetime is 500 days and the standard deviation is 61 days. To the nearest percentage, what percentage of batteries have lifetimes longer than 561 days?

16% of batteries have lifetimes longer than 561 days.

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:

To the nearest percentage, what percentage of batteries have lifetimes longer than 561 days?

This is 1 subtracted by the pvalue of Z when X = 561. So

has a pvalue of 0.8413.

1-0.8413 = 0.1587

Rounding to the nearest percentage, 16% of batteries have lifetimes longer than 561 days.

could you reply to this with the question so i can ?

your answer is a

step-by-step explanation:

4*4*4*4=256