Dmop Homework 4 (section E)

DMOP Homework 4 (Section E)

Q1

i) A: event when painting process is defective

B: event when polishing process is defective

P(A) = 0.2                P(B) = 0.1

Probability of car returning to special shop is P(C) = P(A)*(1-P(B)) + (1-P(A))*P(B) + P(A)*P(B)

                                                 = 0.2*0.9 + 0.8*0.1 + 0.2*0.1

                                                 = 0.28

ii) Expected value of C = n*P(X)

                           = 1000*0.28

                           = 280

iii) Standard deviation (C) = SQRT(n*p(1-p))

                           14.19

For X = 200 the value of z = (200-280)/14.19

                            =-5.53

P(Z<=z) ~ 0

the probability that the number of returned cars is less than or equal to 200 is almost zero

iv) X: binomial distribution

     Y: binomial distribution

v) X = 200

    Y = 100

X+Y = 300

z = (300-300)/SD

   = 0

P(Z<=z) = 0.5

Q2. N = 300                sample mean = $739.98                 

Sample standard deviation = $312.7

a) 95% confidence interval

[739.98 + 1.96*(312.7/ SQRT(300)), 739.98 - 1.96*(312.7/SQRT(300))]

[775.36, 704.59]

b) 99% confidence interval

 [739.98 + 2.576*(312.7/ SQRT(300)), 739.98 – 2.576*(312.7/SQRT(300))]

   [786.48, 693.47]

c) 30 = 1.96*312.7/ SQRT(n)

   n = 418

Q3

[pic 1]

Mean of annual net income of ABC = $9961.21

We take 10,000 iterations for this simulation

Standard deviation = 1319.25

Mean standard error = 1319.25/sqrt(10,000)

                          = 13.19

95% confidence interval

(9961.21+ 1.96 * 13.19, 9961.21 - 1.96 * 13.19)

(9987.06, 9935.35)

Q4

Expected Profit = 0.6* Profit of hit + 0.4 * profit of non hit

= 46000

[pic 2]

The risk of company losing money is 20.7% here.

Q5

a) Probability of Strategy A being less profitable than Strategy B is between 0.3 and 0.35 ~ 0.31

The probability of  the profit from Strategy A shall be more than that of B is almost equal to 0.69