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:

Answer:

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

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

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 [tex]\mu = 30, \sigma = 2[/tex]

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

So [tex]n = 400, s = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{400}} = 0.1[/tex]

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

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