Prim 036 exam pack

PRIM036 EXAM PACK 2018
MAY JUNE 2018
QUESTION 1
⋏𝑥 .𝑒 −⋏
a. P (X = x ) = 𝑥!
⋏=3
= P (x ≤ 3) = P(x= 0) + P(x=1) +P (x=2) + P (x=3)
30 .𝑒 −3 31 .𝑒 −3 32 .𝑒 −3 33 .𝑒 −3
= + + +
0! 1! 2! 3!


= 0.0498 + 0.1494 + 0.2240 + 0.2240
= 0.6472
3
b. ⋏ = 3 =1
P (x ≥ 2) = 1- P(x= 0)- P(x=1)
10 .𝑒 −1 11 .𝑒 −1
=1- -
0! 1!
= 1- 0.3679 -0.3679
= 0.2642




QUESTION 2
SIZE POOR AVERAGE GOOD MINIMUM IN
A ROW
SMALL 2 6 9 2

MEDIUM 3 5 10 3

LARGE 2 3 11 2




Maximum of the minimums
i. Maximin = 3 which corresponds with medium

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, ii. Maximax

SIZE POOR AVERAGE GOOD MAXIMUM
INA ROW
SMALL 2 6 9 9

MEDIUM 3 5 10 10

LARGE 2 3 11 11




Maximum
Maximax = 11,which corresponds with large




iii. Minimax

SIZE POOR AVERAGE GOOD MAXIMUM IN
A ROW
SMALL (3-2) =1 (6-6)= 0 (11-9)=2 2

MEDIUM (3-3)= 0 (6-5)=1 (11-10)=1 1

LARGE (3-2) =1 (6-3)=3 (11-11)=0 3



Minimum in a column
Minimax = 1,which corresponds with medium




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CONTACT : 062 914 2414

, b. EMV =∑pi x mi
PAYOFF TABLE

SIZE POOR AVERAGE GOOD EMV

SMALL 2 6 9 (2x0.15) +
(6x0.45)+(9x0.4)= 6.6
MEDIUM 3 5 10 (3x0.15) +
(5x0.45)+(10x0.4)= 6.7
LARGE 2 3 11 (2x0.15) +
(3x0.45)+(11x0.4)= 6.05
PROBABILITY 0.15 0.45 0.4



The manufacturer have to choose the medium which has an EMV of 6.7 (670 000)




c. EOL =∑pi x li



SIZE POOR AVERAGE GOOD EMV

SMALL 1 0 2 (1x0.1) +
(0x0.45)+(2x0.45)= 1
MEDIUM 0 1 1 (0x0.1) +
(1x0.45)+(1x0.45)= 0.9
LARGE 1 3 0 (1x0.1) +
(3x0.45)+(0x0.45)= 1.45
PROBABILITY 0.1 0.45 0.45



Yes he should accept the offer because the minimum EOL is R90000 (0.9) which is more
than 25 000 offered.
3
TUTOR :ADDIE
CONTACT : 062 914 2414