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The batteries produced in a manufacturing plant have a mean time to failure of 30 months with a standard deviation of 2 months. I select a simple random sample of 400 batteries produced in the manufacturing plant. I test each and record how long it takes for each battery to fail. I then compute the average of all the failure times. The sampling distribution of is approximately:

User MikeyB
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Answer:

By the Central Limit Theorem, the sampling distribution is approximately normal, with mean 30 and standard deviation 0.1.

Explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
\mu and standard deviation
\sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
\mu and standard deviation
s = (\sigma)/(√(n)).

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
\mu = p and standard deviation
s = \sqrt{(p(1-p))/(n)}

The batteries produced in a manufacturing plant have a mean time to failure of 30 months with a standard deviation of 2 months.

This means that
\mu = 30, \sigma = 2

Sample of 400 batteries. The sampling distribution of is approximately:

So
n = 400, s = (\sigma)/(√(n)) = (2)/(√(400)) = 0.1

By the Central Limit Theorem, the sampling distribution is approximately normal, with mean 30 and standard deviation 0.1.

User Karol S
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