# Decision Analysis

**Topics:**Probability theory, Variance, Normal distribution

**Pages:**7 (438 words)

**Published:**July 14, 2014

Bill Denomination (X)

Number of Bills (f)

X*f

X2*f

$1

520

520

520

$5

260

1,300

6500

$10

120

1,200

12000

$20

70

1,400

28000

$50

29

1,450

72500

$100

1

100

10000

Total

1000

5,970

129520

a.

b.

c. The probability that a bar containing $50 or $100 bill is purchased is 30 / 1000 = 0.03 Hence customer have to buy 100 bars of soap, so that he or she has purchased three bars containing a $50 or $100 bill.

d. Given n=1000, E(X) = 5.97, SD(x) = 9.68912

The probability that a soap contains a bill of $20 or above is 100 / 1000 = 0.0.1 Binomial with n =8, p = 0.1

Hence the probability that at least one of these bar contains a bill of $20 or larger is 0.5595

Q.6 Answer

Number of Children

Frequency

1996

2006

1

3671455

3971276

2

100750

137085

3

5298

6118

4

560

355

5

81

67

a.

X

1

2

3

4

5

P (x)

0.97125

0.02665

0.00140

0.00015

0.00002

b.

x

1

2

3

4

5

Total

P(x)

0.97125

0.02665

0.00140

0.00015

0.00002

1.00

x*P (x)

0.97125

0.0533

0.0042

0.00059

0.00011

1.03

x2*P (x)

0.97125

0.10661

0.01261

0.00237

0.00054

1.09

Expected value = E(x) = ∑ x*P (x) = 1.03

Variance = E(x2) – [E(x)]2 = 1.09-1.03*1.03 = 0.0291

c.

y

1

2

3

4

5

Total

P(y)

0.96463

0.03330

0.00149

0.00009

0.00002

1.00

d.

y

1

2

3

4

5

Total

P(y)

0.96463

0.03330

0.00149

0.00009

0.00002

1.00

y*P (y)

0.96463

0.06660

0.00446

0.00034

0.00008

1.04

y2*P (y)

0.96463

0.13319

0.01337

0.00138

0.00041

1.11

Expected value = E(x) = ∑ x*P (x) = 1.04

Variance = E(x2) – [E(x)]2 = 1.11-1.04*1.04 = 0.0284

e. The expected value of the number of children born in a single pregnancy in 1996 was 1.03 and in 2006 it was 1.04. hence we do not support the conclusion that increased use of fertility drugs by older women has generated an upward trend in multiple births. Q.7 Answer

Unit demand (x)

300

400

500

600

Total

p(x)

0.2

0.3

0.35

0.15

1

x* p(x)

60

120

175

90

445

x2* p(x)

18000

48000

87500

54000

207500

a. E(x) = 445,

Hence Carolina’s monthly order quantity should be 445.

b. Total cost of 445 units will be = 445 * 50 = $22250

Total revenue generated on 300 units = 300*70 = $21000

Hence the loss of company in a month will be = 22250-21000 = $1250

c. Variance of number of units demanded = E(x2) – [E(X)]2 = 207500 – 445 * 445

= 9475

SD of number of units demanded = sqrt ( 9475) = 97.3396

Q.11 Answer

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

Binomial with n = 6, p = 0.23

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Binomial with n = 6, p = 0.23

c) In a sample of ten British citizens, what is the probability that none have boycotted good for ethical reasons?

Binomial with n = 10, p = 0.23

Q.25 Answer

Let X: The stock price for companies making u the S&P 500.

Given X follows Normal distribution with mean = 30 and SD = 8.20 a.

b.

c. To find ‘z” such that P(Z >z)=0.1 which implies z = -1.282

Hence ‘A’ = z * 8.20 + 30 = 19.4876

Q.26 Answer

Let X: Age of accounts.

Given X follows Normal distribution with mean = 28 and SD = 8 a. P (20 < x < 40) = P( x < 40) – P (x < 20)

= P (z1< 1.5 ) – P (z2< -1)

= 0.93319 – 0.15866

= 0.77453

b. To find z such that P (Z 7) = P (z > 1.7889) =

=1 – P (z < 1.7889)

= 1-0.96318

= 0.03682

Hence the probability that store will have to refund the money to its December 17 customers is 0.03682

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