STAT217

Worksheet #1 Solutions

Question 1

a) 0.4429 from the table

b) 0.4993 from the table

c) 0.3997 + 0.2486 = 0.6483

d) 0.4945 – 0.4177 = 0.0768

e) 0.4761 – 0.008 = 0.4681

f) We need the 85.08^th percentile. From this we get P(0 < Z < x) = 0.3508. Looking for the probability 0.3508 in the Z table, we find the corresponding Z value to be 1.04

g) Since P(-x < Z < x) = 0.9756, P(0 < Z < x) = 0.4878. As with part f, we find the corresponding Z value to be 2.25.

Question 2

a) Z = (7-5)/2 = 1. Then P(X < 7) = P(Z < 1) = 0.5 + 0.3413 = 0.8413.

b) Z = (9.2-5)/2 = 2.1. Then P(X > 9.2) = P(Z > 2.1) = 0.5 – 0.4821 = 0.0179

c) Z1 = (4-5)/2 = -0.5; Z2 = (6.4-5)/2 = 0.7. Then P(4 < X < 6.4) = P(-0.5 < Z < 0.7) = 0.1915 + 0.258 = 0.4495

d) We need the 93.7^th percentile. From this we get P(0 < Z < x) = 0.437. The Z value is 1.53. Then 1.53 = (x-5)/2. Solving for x, x = 5 + 2(1.53) = 8.06.

e) Since P(-x < Z < x) = 0.9544, then P(0 < Z < x) = 0.4772. The Z value is 2. For the lower limit, we have –2 = (x – 5)/2 from which we get x = 5 – (2)(2) = 1; For the upper limit, we have 2 = (x – 5)/2 from which we get x = 5 + 2(2) = 9

Question 3

a) Z = (10.01-10.2)/0.2 = -0.95. Then P(X < 10.01) = P(Z < -0.95) = 0.5 – 0.3289 = 0.1711

b) Z = (10.3-10.2)/0.2 = 0.5. Then P(X > 10.3) = P(Z > 0.5) = 0.5 – 0.1915 = 0.3085

c) In this case we need the 5^th percentile of Z which is –1.645. Then –1.645 = (x-10.2)/0.2. Solving for x, x = 10.2 – 1.645(0.2) = 9.871 seconds.

Question 4

Z = (4.3 – 4)/(1.2/sqrt(50)) = 1.77.

P(Xbar > 4.3) = P(Z > 1.77) = 0.5 – 0.4616 = 0.0384.

Question 5

Z = (4.3 – 4)/(1.2/sqrt(100)) = 2.5.

P(Xbar > 4.3) = P(Z > 2.5) = 0.5 – 0.4938 = 0.0062.

Question 6

a) Lower limit = 15.14 – 1.96(2.54)/sqrt(100) = 15.14 – 0.50 = 14.64

Upper limit = 15.14 + 0.5 = 15.64.

14.64 < µ 15.64

With 95% confidence, the average amount that people spend per week on fast food ranges from $14.64 to $15.64.

b) The Z value changes from 1.96 to 2.576. So, the margin of error changes to 2.576(2.54)/sqrt(100) = 0.65.

Lower limit = 15.14 – 0.65 = 14.49

Upper limit = 15.14 + 0.65 = 15.79

Since the margin of error is larger, the confidence interval is wider.

c) Once again, the Z value is 1.96. The margin of error is 1.96(2.54)/sqrt(200) = 0.35.

Lower limit = 15.14 – 0.35 = 14.79

Upper limit = 15.14 + 0.35 = 15.49

Since the margin of error is smaller, the confidence interval is narrower.

d) The margin of error is 1.96(5.08)/sqrt(100) = 1.00

Lower limit = 15.14 – 1= 14.14

Upper limit = 15.14 + 1= 16.14

Since the margin of error is larger, the confidence interval is wider.

e) Sample size = n = (1.96(2.54)/0.5)² = 9.96² = 99.2 which we round up to 100.

Question 7

Lower limit = 2650 – 1.96(720)/sqrt(120)sqrt((800-120)/(800 – 1)) = 2650 – 119 = 2531

Upper limit = 2650 + 119 = 2,769

$2,531 < µ < $2,769

Question 8

n = [800(720²)]/[799(50/1.96)² + 720²] = 399.4 which rounds up to 400.

Question 9

Sample size = n = [180(0.85)(0.15)]/[179(0.02/1.96)² + 0.85(0.15)] = 157.04 which we round to 158.

Question 10

E = 1.96sqrt(0.85(0.15)/80)sqrt((180-80)/(180-1)) = 0.0585

Question 11

0.05 = Zsqrt(0.85(0.15)/80)sqrt(100/179) from which we get Z = 1.6757. If we round Z to 1.68, then P(0 < Z < 1.68) = 0.4535. We multiply this by 2 to get the level of confidence of 90.7%

From the textbook

Section 6-2

Question 29 (page 314)

Part a

Lower limit = 132 - 1.96(10)/sqrt(31)sqrt(69/99) = 132 – 2.9389 = 129.0611

Upper limit = 132 + 2.9389 = 134.9389

129.0611 < µ < 134.9389

Part b

n = [200(10²)(1.96²)]/[199(1.5²) + (10²)(1.96²)] = 92.3 which we round up to 93.

Question 32 (page 315)

Parts a, b and c together

Xbar = 450 since it is halfway between the lower and upper limits. This gives E = 20. Then 20 = 1.96s/sqrt(100) from which we get s = 102.0408163. For a 99% confidence interval, Z = 2.575.

Lower limit = 450 – 2.575(102.0408163)/sqrt(100) = 450 – 26.2755 = 423.7245

Upper limit = 450 + 26.2755 = 476.2755

423.7245 < µ < 476.2755

Part d

E = 468 – 450 = 18. Then 18 = Z(102.0408163)/sqrt(100) from which we get Z = 1.764. P(0 < Z < 1.764) is approximately 0.4612 which we multiply by 2 to get a confidence level of 92.24%. If we round Z to 1.76, the approximate level of confidence is 92.16%.

Section 6-4

Question 29 (page 337)

For 97% confidence, we have 1.5% in either tail. Then P(0 < Z < cv) = 0.485 from which we get cv = 2.17. Since p-hat is unknown, we use p-hat = 0.5.

n = [5000(0.25)(2.17²)]/[4999(0.02²) + 0.25(2.17²)] = 1,852.8 which we round up to 1,853.

Question 31 (page 338)

We have 0.002 = Zsqrt((0.08)(0.92)/47,000) from which we get Z = 1.5982 which we can round to 1.60. P(0 < Z < 1.6) = 0.4452 which we multiply by 2 to get the confidence level of 89.04%.