The batteries produced in a manufacturing plant have a mean time to failure of 27 months with a standard deviation of 2 months. a simple random sample of 100 batteries is taken and it is recorded how long it takes for each battery to fail. what is the level l so there is only a 4% chance that the sample mean time to failure falls above l? round your answer to two decimal places.

There is only a 4% chance that the sample mean time to failure falls above 27.35 months.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

For this problem, we have that:

The batteries produced in a manufacturing plant have a mean time to failure of 27 months with a standard deviation of 2 months, so .

We have a sample of 100 students, so we need to find the standard deviation of the sample, to use in the place of in the z score formula.

What is the level L so there is only a 4% chance that the sample mean time to failure falls above L?

This level is the value of X when Z has a pvalue of 0.96.

has a pvalue of 0.96. So we have to find X when .

There is only a 4% chance that the sample mean time to failure falls above 27.35 months.

the answer is (-4,-4)

hope this ! : d

really really sorry if i'm

-12 -5(x + 4) = 3x

-12 - 5x - 20 = 3x

-12 - 20 - 5x = 3x

-32 - 5x = 3x

    + 5x   + 5x

-32 = 8x

8         8

- 4= x