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Solutions to Statistics Problems: X Values, Passengers, and Probabilities - Prof. Shili Li, Assignments of Statistics

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

Pre 2010

Uploaded on 07/23/2009

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koofers-user-4me 🇺🇸

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Hw3-solution
3.6
Possible X values are1, 2, 3, 4, … (all positive integers)
Outcome: RL AL RAARL RRRRL AARRL
X: 2 2 5 5 5
3.10
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
3.12
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 = .83
b. Using the information in a. above, P(Y > 50) = 1 - P(Y 50) = 1 - .83 = .17
c. For you to get on the flight, at most 49 of the ticketed passengers must show up. P(Y
49) = .05 + .10 + .12 + .14 + .25 = .66. For the 3rd person on the standby list, at most 47
of the ticketed passengers must show up. P(Y 47) = .05 + .10 + .12 = .27
3.14
a.
5
1
)(
y
yp = K[1 + 2 + 3 + 4 + 5] = 15K = 1 15
1
K
b. P(Y 3) = p(1) + p(2) + p(3) = 4.
15
6
c. P( 2 Y 4) = p(2) + p(3) + p(4) = 6.
15
9
d. 1
50
55
]2516941[
50
1
50
5
1
2
y
y; No
pf2

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Hw3-solution

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 ≤

  1. = .05 + .10 + .12 + .14 + .25 = .66. For the 3rd^ person on the standby list, at most 47 of the ticketed passengers must show up. P(Y ≤ 47) = .05 + .10 + .12 =.

a. 

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

[ 1 4 9 16 25 ]^55

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