An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. He believes that the mean income is $24.8, and the variance is known to be $125.44. How large of a sample would be required in order to estimate the mean per capita income at the 85% level of confidence with an error of at most $0.59

747 samples

Step-by-step explanation:

Given:

Standard deviation = √125.44 = 11.2

Zcritical = 85℅

Margin of error, E = 0.59

The sample size, n required cnanbe obtained using the relation :

n = [(Zα/2 * σ) / E]²

Zcritical at 85% = 1.44

n = [(1.44 * 11.2) / 0.59]²

n = (16.128 / 0.59)²

n = 747.23

n = 747 samples

10 glasses of juice could be made with 5 carrots, 1,500 orange juice, 2,000 papaya juice.

78.59 is the mean

step-by-step explanation: