Suppose a computer engineer is interested in determining the average weight of a motherboard manufactured by a certain company. A summary of a large sample provided to the engineer suggest a mean weight of 11.8 ounces and an estimated standard deviation, sigma = 0.75. How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4? How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,

Step-by-step explanation:

How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4?

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error(width) as such

In which is the standard deviation of the population and n is the size of the sample.

For this item, we have:

. So

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error(width) as such

In which is the standard deviation of the population and n is the size of the sample.

For this item, we have:

. So

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,

ok thats very interesting

vertical asymptote .

horizontal asymptote .

step-by-step explanation:

given function is .

vertical asymptote is given by setting

denominator =0

when degree of numerator and denominator are same then divide first term of numerator by first term of denominator to get horizontal asymptote.