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Statistics Homework Solutions for Stat 400, Assignments of Probability and Statistics

University of MarylandProbability and Statistics

Solutions to homework problems for a statistics 400 course, covering topics such as probability mass functions, expected values, and variance. It includes calculations for poisson distribution, binomial distribution, and normal distribution.

Typology: Assignments

2019/2020

Uploaded on 11/25/2020

koofers-user-awg
koofers-user-awg 🇺🇸

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Stat 400 HW 2 Solution
TA: Emily King
1. Let Y= # forms required of next applicant, p(y) = ky, y = 1, . . . , 5
(a) Since 1 = P5
y=1 p(y) = P5
y=1 ky =k5(5+1)
2= 15k,k=1
15 .
(b)
P(X≤3) = 1 −P(X≥4)
= 1 −(4
15 +5
15)
=6
15
=2
5
(c)
P(2 ≤X≤4) = 2
15 +3
15 +4
15
=9
15
=3
5
(d) In order to be a pmf, P5
y=1 p(y) = 1. If p(y) = y2
50 for y= 1, . . . , 5,
then
5
X
y=1
y2
50 =1
50
5
X
y=1
y2
=1
50 5·6·11
6
=55
50
6= 1
pf3
pf4

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Download Statistics Homework Solutions for Stat 400 and more Assignments Probability and Statistics in PDF only on Docsity!

Stat 400 HW 2 Solution

TA: Emily King

  1. Let Y = # forms required of next applicant, p(y) = ky, y = 1,... , 5

(a) Since 1 =

5

y=

p(y) =

5

y=

ky = k

5(5+1)

2

= 15k, k =

1

15

(b)

P (X ≤ 3) = 1 − P (X ≥ 4)

(c)

P (2 ≤ X ≤ 4) =

(d) In order to be a pmf,

5

y=

p(y) = 1. If p(y) =

y

2

50

for y = 1,... , 5,

then

5 ∑

y=

y

2

5 ∑

y=

y

2

  1. (a)

p(m) =

1 / 36 m = 1

3 /36 = 1/ 12 m = 2

5 / 36 m = 3

7 / 36 m = 4

9 /36 = 1/ 4 m = 5

11 / 36 m = 6

0 otherwise

(b)

P (M ≤ m) =

0 m < 1

1 / 36 1 ≤ m < 2

4 /36 = 1/ 9 2 ≤ m < 3

9 /36 = 1/ 4 3 ≤ m < 4

11 /36 = 4/ 9 4 ≤ m < 5

25 / 36 5 ≤ m < 6

1 m ≥ 6

  1. E(X) = 1(.2) + 2(.4) + 3(.3) + 4(.1) = 2.3 lots

E(X

2 ) = 1(.2) + 4(.4) + 9(.3) + 16(.1) = 6. 1

V (X) = E(X

2 ) − [E(X)]

2 =. 81

Let Y = rv which is the # of lbs left after the next customer’s order is

shipped. Then E(Y ) = E(100 − 5 X) = 100 − 5 E(X) = 88.5 lbs, and

V (Y ) = 25V (X) = 20.25.

  1. 25% of incoming calls involve fax messages. (p = .25)

Take a sample of 25 incoming calls. (n = 25).

This is clearly modeled by a Binomial distribution.

(a) P (X ≤ 6) = B(6; 25, .25) ≈. 561

(b) P (X = 6) = B(6; 25, .25) − B(5; 25, .25) ≈. 183

(c) P (X ≥ 6) = 1 − B(5; 25, .25) ≈. 622

(e) Let Y = time spent taking exam. Then E(Y ) = 3(.98) + 4.5(0.2) =

3 .03 hours.