An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.

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Answer:

The 96% confidence interval for the population mean of all bulbs produced by this firm is between 765 hours and 795 hours.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.96}{2} = 0.02[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex]

So it is z with a pvalue of 1-0.02 = 0.98, so z = 2.055

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

So

[tex]M = 2.055*\frac{40}{\sqrt{30}} = 15[/tex]

The lower end of the interval is the mean subtracted by M. So 780 - 15 = 765 hours.

The upper end of the interval is M added to the mean. So 780 + 15 = 795 hours.

The 96% confidence interval for the population mean of all bulbs produced by this firm is between 765 hours and 795 hours.

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