The ages of a group of 142 randomly selected adult females have a standard deviation of 18.1 years. Assume that the ages of female statistics students have less variation than ages of females in the general​ population, so let σ=18.1 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics​ students? Assume that we want 99​% confidence that the sample mean is within​ one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general​ population? The required sample size is

Start with the formula for Z:

Z = (x-µ)/(σ/√n)

We want the sample mean to be within one-half year of the population mean, so we set x-µ=0.5. We are looking for a 99% confidence interval, so we set Z=2.7578. We are told to use σ=18.1. Plugging those values into the formula, we get:

2.5758 = 0.5(18.1/√n)

We can rearrange to solve for n:

((2.5758-18.1)/0.5)2 = n

Plugging that into our calculator, we get n = 964.003. Since we can't have a fraction of a person in our sample, it would be safest to round up to n=965. (But since .003 is so small, I'd also accept 964 as an answer.)Step-by-step explanation:

all you should have to do is insert 1 where the c is, representing that one car is sold.

f(c) = 250c + 500

f(1) = 250(1) + 500

f(1) = 250 + 500

f(1) = 750

he earns $750 for each car sold.

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