Suppose you believe that the true average daily trade volume for General Electric stock is 49,829,719 shares and a standard deviation of 21,059,637 shares. Considering a 95% confidence level: What is the minimum required sample size if you would like your sampling error to be limited to 1,000,000 shares

The minimum sample size is 1,704.

Step-by-step explanation:

We have 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 Z-table as such z has a p-value of .

That is z with a p-value of , so Z = 1.96.

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.

Standard deviation of 21,059,637 shares

This means that

What is the minimum required sample size if you would like your sampling error to be limited to 1,000,000 shares?

This is n for which , so:

Rounding up:

The minimum sample size is 1,704.

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)