A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 3364 with a mean life of 530 minutes. If the claim is true, in a sample of 75 batteries, what is the probability that the mean battery life would be greater than 533.2 minutes? Round your answer to four decimal places.

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a sample mean. The distribution for sample means follows a normal distribution with mean and standard deviation . Mathematically

For values of the sample , no matter the distribution the data come from.

And the variable z follows a standard normal distribution, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

[1]

That is

We have a variance of 3364. That is, a standard deviation of

The population mean is

The sample size is

The sample mean is

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

With this value of z we can consult a cumulative standard normal table (or use some statistic program) to find the cumulative probability for z (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

However, in the question we are asked for . As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

Thus

Rounding to four decimal places, we have

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) .

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the answer is d on edge

step-by-step explanation: