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Solutions to various statistics problems involving determining possible x values, passenger capacities, and probabilities. Topics include calculating probabilities of certain events, finding the probability of x being less than a certain value, and determining the probability of x being between two values.
Typology: Assignments
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Possible X values are1, 2, 3, 4, … (all positive integers)
Outcome: RL AL RAARL RRRRL AARRL X: 2 2 5 5 5
a. T = total number of pumps in use at both stations. Possible values: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
b. X: -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6
c. U: 0, 1, 2, 3, 4, 5, 6
d. Z: 0, 1, 2
a. In order for the flight to accommodate all the ticketed passengers who show up, no more than 50 can show up. We need Y ≤ 50. P(Y ≤ 50) = .05 + .10 + .12 + .14 + .25 + .17 =.
b. Using the information in a. above, P(Y > 50) = 1 - P(Y ≤ 50) = 1 - .83 =.
c. For you to get on the flight, at most 49 of the ticketed passengers must show up. P(Y ≤
5
1
y
p y = K[1 + 2 + 3 + 4 + 5] = 15K = 1 K 151
b. P(Y 3) = p(1) + p(2) + p(3) = 156 . 4
c. P( 2 Y 4) = p(2) + p(3) + p(4) = 159 . 6
d. 1 50
5
1
2 ^
y
y (^) ; No
a. p(1) = P(M = 1) = P[(1,1)] = 361
p(2) = P(M = 2) = P[(1,2) or (2,1) or (2,2)] = 363
p(3) = P(M = 3) = P[(1,3) or (2,3) or (3,1) or (3,2) or (3,3)] = 365
Similarly, p(4) = 367 , p(5) = 369 , and p(6) = 3611
b.
F(m) =
36 25 36 16 36 9 36 4
361
m
m
m
m
m
m
m
a. Possible X values are those values at which F(x) jumps, and the probability of any particular value is the size of the jump at that value. Thus we have:
x 1 3 4 6 12
p(x) .30 .10 .05 .15.
b. P(3 X 6) = F(6) – F(3-) = .60 - .30 =. P(4 X) = 1 – P(X < 4) = 1 – F(4-) = 1 - .40 =.
0 1 2 3 4 5 6 7