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Westkost [7]
3 years ago
13

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:
Mathematics
1 answer:
kompoz [17]3 years ago
6 0

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 \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 = \frac{\sigma}{\sqrt{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{\frac{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 = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{400}} = 0.1

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

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