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