Stat 118 - Chapter 6 Homework and Excel Lab

Stat 118 - Chapter 6 Homework and Excel Lab


1. The U.S. Postal Service reports 95% of first-class mail within the same city is delivered within 2 days of the
time of mailing. Six letters are randomly sent to different locations.

a. What is the probability that all six arrive within 2 days? (Round your answer to 3 decimal places.)

b. What is the probability that exactly five arrive within 2 days? (Round your answer to 3 decimal places.)

c. Find the mean number of letters that will arrive within 2 days. (Round your answer to 1 decimal place.)
d-1. Compute the variance of the number that will arrive within 2 days. (Round your answer to 3 decimal places.)

d-2. Compute the standard deviation of the number that will arrive within 2 days. (Round your answer to 3
decimal places.): n = 6
p = 0.95
X = number of mails arrive within two days

a. 0.735
P(X = 6) = (0.95)^6 = 0.7351

b. 0.232
P(X = 5) = 6 x (0.95)^5 x (1-0.95)^(6-5)
= 6 x 0.7737809375 x 0.05^1
= 0.232

c. 5.7
MEAN:
E(X) = np
6 x 0.95 = 5.7

d-1. 0.285
VARIANCE: V(X) =
np(1-p)
= 6 x 0.95 x (1-0.95)
= 6 x 0.95 x 0.05




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= 0.285

d-2. 0.534
STANDARD DEVIATION:
à =square root of V(X)
Square root of 0.285 = 0.5338
2. [PART ONE]
In a recent study, 90% of the homes in the United States were found to have large-screen TVs. In a sample of
nine homes, what is the probability that:

a. All nine have large-screen TVs? (Round your answer to 3 decimal places.)

b. Less than five have large-screen TVs? (Round your answer to 3 decimal places.): n = 9
p = 0.90
P(X) = (combination formula) x (p^n) x ((n-p)^n-x)

a) 0.387
P(X = 9) = 0.90 ^ 9 = 0.387

b) 0.001
nCr = MATH --> PRB --> 3:nCr

P(X = 0) = 9C0 x 0.90^0 x (1 - 0.90)^9-0 P(0) = 1 x 0.90^0 x
0.10^9
= 0.000000001

P(X = 1) = 9C1 x 0.90^1 x (1 - 0.90)^9-1 P(1) = 9 x 0.90^1 x
0.10^8
= 0.000000081

P(X = 2) = 9C2 x 0.90^2 x (1 - 0.90)^9-2 P(2) = 36 x 0.90^2 x
0.10^7
= 0.000002916

P(X = 3) = 9C3 x 0.90^3 x (1 - 0.90)^9-3 P(3) = 84 x 0.90^3 x
0.10^6
= 0.000061236




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P(X = 4) = 9C4 x 0.90^4 x (1 - 0.90)^9-4 P(4) = 126 x 0.6561 x
0.10^5
= 0.000826686

P(X < 5 ) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)
0.000000001 + 0.000000081 + 0.000002916 + 0.000061236 + 0.000826686

= 0.00089092
3. [PART TWO]
In a recent study, 90% of the homes in the United States were found to have large-screen TVs. In a sample of
nine homes, what is the probability that:

c. More than five have large-screen TVs? (Round your answer to 3 decimal places.)

d. At least seven homes have large-screen TVs? (Round your answer to 3 decimal places.): c) 0.992
Using the complement rule, P(A) = 1 - P(~A), P(X>5) = 1 - P(X <= 5)
1 - [(P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)]
1 - [0.00089092 + P(X=5)

P(X = 5) = 9C5 x 0.90^5 x (1 - 0.90)^9-5
= 0.007440174

= 0.00089092 + 0.007440174
=0.008331094

1 - 0.008331094
= 0.992

d) 0.947
P(X>=7) = P(X=7) + P(X=8) + P(X=9)
(9C7 x 0.90^7 x 0.10^2) + (9C8 x 0.90^8 x 0.10^1) + (9C9 x 0.90^9 x 0.10^0)
= 0.947
4. The speed with which utility companies can resolve problems is very im- portant. GTC, the Georgetown
Telephone Company, reports it can resolve customer problems the same day they are reported in 70% of the
cases. Suppose the 15 cases reported today are representative of all complaints.




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