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Business Statistics (BUS 505) Assignment 6

Answer to the question no -34

Let, Ai = Frequency of Unit [i = 1, 2, 3….] Bj = Purchase of generic products [j = 1, 2, 3….]

a. P(A1¿ B1 )= P(A1) P(B1) = .19¿ .79 = .12

b. ( A1│B3) =

P (A1∩B3 )P(B3) =

. 19

. 27 =.7037

c. ( A1│B3) = P(A1) P(B3) .19 = .79¿ .27 .19 ≠ .2133So, they are not statistical independent.

d. (B1│A2) =

P (B1∩A1 )P (A2 ) =

. 07

.21 =.333 e. P(B1¿ A2)=P(B1) P(A1) .07 = .19¿ .21 .07 ≠ .0399 So, they are not statistical independent.f. P(A1) = P(B1)+P(B2)+P(B3) =.12+.48+.19 =.79g. P (B3) = P (A1) +P (A2) = .19+.08 =.27h. P (A1¿ B3 )=P(A1)+P(B3)- P(A1¿ B3) = .79+.27-.19 =.87

FREQUENCY OF VISIT

PURCHASE OF GENERIC PRODUCT

OFTEN(B1) SOMETIMES(B2) NEVER(B3) TOTAL

Frequent(A1) .12 .48 .19 .79

Infrequent(A2) .07 .06 .08 .21

TOTAL .19 .54 .27 1.00

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Business Statistics (BUS 505) Assignment 6

Answer to the question no -36TradedStock

READ BUSSINESS SECTION

Regular (B1) Occasional(B2) Never (B3) Total

Yes (A1) .18 .10 .04 .32

No (A2) .16 .31 .21 .68

Total .34 .41 .25 1.00

Let, Ai =Traded Stock [i = 1, 2, 3….] Bj =Read Business section [j = 1, 2, 3….]

a. P (B3) =P (A1) +P (A2) =.04+.21 =.25

b. P (A1) =.18+.10+.04 =.32

c. P (A1│B3) =

P (A 1∩B2 )P(B2) =

. 04

. 25 =.16

d. P (B3│A1) =

P (B3∩A 1)P(A1 ) =

. 04

. 32 =.125 e. Does not read regularly= P (B1 ) = 1- P (B1) =1-.34 = .66

P (A1│B1 ) =

P (A1∩B1 )P(B1)

We Know, P (A1) =P (A1¿ B1) +P (A1¿ B1 ) .32 =.18 + P (A1¿ B1 ) P (A1¿ B1 ) = .32-.18 =.14

P (A1¿ B1 ) =

. 14

. 66 =.212

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Business Statistics (BUS 505) Assignment 6

Answer to the question no -37

PART SUBCONTRACTORA(B1) B(B2) C(B3) Total

Good(A1) .27 .30 .33 .90

Defective (A2) .02 .05 .03 .10Total .29 .35 .36 1.00

Let, Ai=Particular Sensitive part [i = 1, 2, 3….] Bj =Subcontractor [j = 1, 2, 3….]

a. P(A2)= .02+.05+.03 =.10b. P (B2) = .30+.05 = .35

c. P (A2│B2) =

P (A 2∩B2 )P(B2) =

. 05

.35 =.1428

d. P(B2│A2)=

P (B2∩A2 )P (A2 ) =

. 05

. 10 =.5e. P(A1¿ B2)=P(A1)P(B1) = .90¿ .29 .27≠ .261 So, they are not Statistical independent.

f. P(A│G)=

P (A∩G )P (G ) =

. 27

. 9 =.30

P (B│G) =

P (B∩G)P(G) =

. 30

. 9 =.33

P (C│G) =

P (C∩G)P(G ) =

. 33

. 9 =.37∴ In terms of quality subcontractor C is most reliable.

Answer to the question no -38

WorkedProblem

EXPECTED GRADEA(B1) B(B2) C(B3) Below C(B4) Total

YES (A1) .12 .06 .12 .02 .32NO (A2) .13 .21 .26 .08 .68

Total .25 .27 .38 .10 1.00

Let, Ai = Workers related problem [i = 1, 2, 3….]

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Business Statistics (BUS 505) Assignment 6

Bj = Expected grade [j = 1, 2, 3….]

a. P (A1) =.12+.06+.12+.02 = .32

b. P (B1) =.12+.13=.25

c. P (B1│A1) =

P (B1∩A1 )P (A1 ) =

. 12

.32 =.375

d. P (A1│B1) =

P (A1∩B1 )P(B1) =

. 12

.25 =.48

e. Let, R Denotes the Event “Students who worked additional hours expected a Grade below A”

∴P (R│Y) =

P (R∩Y )P(Y ) [P(R) =P(C¿ Y) +P(X<C¿ Y)

=

. 14

. 32 =.12+.02 =.14 ∴P(R¿ Y) =.14

f. P (A1¿ B1) =P (A1) P (B1) .12 = .32¿ .25 .12 ≠ .08 So, they are not statistically independent.

Answer to the question no -41

Let, A denotes the event “A campus student club member of men who receive the material.” B denotes the event “A campus student club member of women who receive the material.” C denotes the event “A campus student club member of student who joins the club.”

So, P (A) = .40 P (B) = .60 P(C∩A )= .07 and P(C∩B)= .09

a) P(C) = P(C∩A )+ P(C∩B) = .07 + .09 = .16

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Business Statistics (BUS 505) Assignment 6

b)P(B/C )=

P(B∩C )P(C ) =

. 09

.16 = .5625

Answer to the question no -42

Let,A1 denotes the event, “Interest rate in next year will be more than 1% than this year.”A2 denotes the event, “Interest rate in next year will be more than 1% lower than this year.”A3 denotes the event, “Interest rates next year within 1% than this year.” B denotes the event, “Significant earnings growth in the next year.”

So, P (A1 ) = .25 P (A2 ) = .15 P(A3 ) = 1-[ P (A1 )+P (A2 )] = .60 ¿

P (B/A1 ) = .10 P(B/A2 ) = .80 P(B/A3 ) = .50

a) P (A1¿ B) = P(B/A1 ) P (A1 ) = .10¿ .25 = .025

b) P(B) = P (A1¿ B) + P(A2¿ B) + (A3¿ B)

= P(B/A1 ) P (A1 ) + P(B/A2 ) P (A2 ) + P(B/A3 ) P(A3 ) = .1¿ .25 + .8¿ .15 + .5¿ .60 = .445

c) P (A2 /B) =

P (A2∩B )P(B) =

P (B/A2 )P (A2 )P(B) =

. 12

. 445 = .270

Answer to the question no -43

Let,A denotes the event, “Blue-collar employees modified health care plan.”B denotes the event, “Blue-collar employees a proposal to change the work schedule.”

P (A) = .42 P(B) = .22 and P(B/A) = .34

a) P (A¿ B) = P(B/A) P (A) = .34 ¿ .42 = .1428b) P(A¿ B) = P(A) + P(B) - P (A¿ B)

= .42 + .22 - .1428 = .4972

c) P(A/B) =

P (A∩B )P (B ) =

. 1428

. 22 = .649

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Business Statistics (BUS 505) Assignment 6