Question

SHOW ALL WORK IN EXCEL

For this assignment, submit one Excel file.

Part 1 sheet is kept blank. Use Part 1 sheet to answer question 1. Put a box around your answer. Round your answer to 4 decimal places.

The Statistical Abstract of the United States published by the U.S. Census Bureau reports that the average annual consumption of fresh fruit per person is 99.9 pounds. The standard deviation of fresh fruit consumption is about 30 pounds. Suppose a researcher took a random sample of 38 people and had them keep a record of the fresh fruit they ate for one year.

What is the probability that the sample average would be less than 90 pounds?

What is the probability that the sample average would be between 98 and 105 pounds?

What is the probability that the sample average would be less than 112 pounds?

What is the probability that the sample average would be more than 93 pounds?

  

[1+ 1.5 + 1 + 1.5 = 5 points]

[Note: Similar problems are available under Lec -1 folder. Solution is also available in the same folder.]

2.

Consider the frame of 40 families with annual incomes shown in sheet named Random Sampling. Generate a simple random sample of size 10 from this population. Calculate the population mean of Annual Income. Then calculate the sample mean of Annual Income. What is the sampling error?

Population
Family	Income
1	$43,300
2	$44,300
3	$34,600
4	$38,000
5	$44,700
6	$45,600
7	$42,700
8	$36,900
9	$38,400
10	$33,700
11	$44,100
12	$51,500
13	$35,900
14	$35,600
15	$43,000
16	$38,600
17	$32,400
18	$22,900
19	$48,100
20	$31,900
21	$56,400
22	$33,600
23	$38,100
24	$42,500
25	$44,900
26	$35,200
27	$60,800
28	$42,500
29	$47,600
30	$36,100
31	$33,000
32	$31,700
33	$48,600
34	$39,300
35	$33,000
36	$36,300
37	$28,400
38	$46,900
39	$37,300
40	$41,000

Answer 1

Part a)

Given :- Population mean   = 99.9 pounds

Standard Deviation = 30 pounds

sample size = 38

Let X :- Average fruit consumption per personfor one year

Using central limit theorem  

P( X < 90) = P ( < (90 - 99.9) / (30/ )

P( X < 90) = P( Z < - 2.03)

P( X < 90) = 0.02118

part b) P ( 98 < X < 105)

Using central limit theorem  

P ( (98 - 99.9)/(30/ ) <   < ( 105 - 99.9) / (30/ )

P ( -0.39< Z < 1.05 )

P ( Z < 1.05) - P ( Z < -0.39)

0.85314 - 0.34827

P ( 98 < X < 105) = 0.50487

Part c)

P ( X < 112) = P ( < ( 112 - 99.9)/ (30/ )

P ( X < 112) = P ( Z < 2.47)

P ( X < 112) = 0.99324

Part d)

P ( X > 93) = P ( > ( 93 - 99.9) / 30/ )

P ( X > 93) = P ( Z > -1.42)

P ( X > 93) = 1 - P ( Z < -1.42)

P ( X > 93) = 1 - 0.07780

P ( X > 93) = 0.9222