Question

Suppose the mean income of firms in the industry for a year is 85 million dollars with a standard deviation of 15 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 107 million dollars? Round your answer to four decimal places.

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. Suppose that the mean income is found to be $21.4 for a random sample of 744 people. Assume the population standard deviation is known to be $5.7. Construct the 95% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

what is the Lower Endpoint?

what is the Upper Endpoint?

Answer 1

Solution :

Given that ,

P(x < 107) = P[(x - ) / < (107 - 85) / 15]

= P(z < 1.47)

= 0.9292

Probability = 0.9292

Sample size = n = 744

Z/2 = 1.96

Margin of error = E = Z/2* ( /n)

= 1.96 * (5.7 / 744)

Margin of error = E = 0.4

At 95% confidence interval estimate of the population mean is,

- E < < + E

21.4 - 0.4 < < 21.4 + 0.4

21 < < 21.8

Lower Endpoint : 21

Upper Endpoint: 21.8