The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 20-inch pizzas. She takes a random sample of 30 pizzas and records their mean and standard deviation as 20.50 inches and 2.10 inches, respectively. She subsequently computes the 90% confidence interval of the mean size of all pizzas as [19.87, 21.13]. However, she finds this interval to be too broad to implement quality control and decides to reestimate the mean based on a bigger sample. Using the standard deviation estimate of 2.10 from her earlier analysis, how large a sample must she take if she wants the margin of error to be under 0.5 inch? (You may find it useful to reference the z table. Round intermediate calculations to at least 4 decimal places and "z" value to 3 decimal places. Round up your answer to the nearest whole number.)

The sample must be of at least 48 pizzas.

Step-by-step explanation:

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 M as such

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

How large a sample must she take if she wants the margin of error to be under 0.5 inch?

She needs a sample of at least n, in which is found when

Rounding up

The sample must be of at least 48 pizzas.

let soccer = x and golf = y

part a

1st equation: x + y = 125

2nd equation: x = y +45

part b

replace x in the 1st equation with the 2nd one:

y + 45 + y = 125

simplify:

2y + 45 = 125

subtract 45 from each side:

2y = 80

divide both sides by 2:

y = 80/2

y = 40

she plays golf for 40 minutes.

part c

no because she plays soccer for 45 minutes longer than golf 40 +45 =   85 minutes, not 80 minutes.

answer: c i think

step-by-step explanation: