Indian Institute of Management Bangalore
PGP 2009-10
Quantitative Methods I
Mid Term Examination

Time: 2 hours 30 minutes Name:____________________________ Maximum Marks: 50 Roll. No.________________ Section____

Question No. | 1 | 2 | 3 | 4 | 5 | Total | Maximum Marks | 3+2+2+2+4=13 | 2+3+4=9 | 2+2*1+2*1.5+4=11 | 3+6+2=11 | 3+3=6 | 50 | Student’s Score | | | | | | |

Instructions:
This is an open-book (1 text-book), open note exam; however you are not allowed to share material with other students. Use of calculator is permitted, but not computer (laptop). Please do not seek any clarifications.

To get any credit, you must

* circle/clearly indicate your final answer (in the space, whenever provided);

* answer all questions in the space provided,

* State any assumptions that you make. Assumptions made should be reasonable.

* Show calculations and provide reasons to support your answers.

Do not attach any additional sheets; use the back sides, if necessary.

1. A highway restaurant is trying to plan its capacity. It finds that on average during the peak lunch hour, which is between 1pm and 2 pm, vehicles arrive at a rate of 1 vehicle per 6 minutes. (Each vehicle on average has four customers, and they can all sit on one table.) To avoid incurring a loss on a particular day, there should be at least 6 vehicles arriving during the peak lunch hour period.

(a) What is the probability of the restaurant incurring a loss on any given day? (3 marks)
X = number of vehicles arriving during peak lunch hour
X ~Poisson (λ = 10)
P [incurring a loss] = P [X < 6] = 0.0671

x | Prob | 0 | 4.53999E-05 | 1 | 0.000453999 | 2 | 0.002269996 | 3 | 0.007566655 | 4 | 0.018916637 | 5 | 0.037833275 | sum | 0.067085963 |

(b) If more than 3 vehicles arrive during a ten minute period, there will be overcrowding and the customers will either have to wait or may decide to leave. What is the probability of this occurring on any given ten minute period? (2 marks)

Y = number of vehicles arriving during a 10-minute period
Y ~Poisson (λ = 5/3)
P [overcrowding] = P [Y > 3] = 1 – P [Y = 0] – P [Y = 1] – P [Y = 2] – P [Y = 3] = 0.0883

y | Prob | 0 | 0.188876 | 1 | 0.314793 | 2 | 0.262327 | 3 | 0.145737 | sum | 0.911733 | 1 - sum | 0.088267 |

(c) They want to tackle the problem of overcrowding. The Manager feels that the peak hour period should be divided into 10 minute intervals. If there are 3 or fewer vehicle arrivals in any specific interval, there is no overcrowding in that interval. What is the probability that there is overcrowding in exactly one such 10 minute intervals within this peak hour period? (2 marks)

Z = number of overcrowded intervals (of 10-minute each)
Z~ Bin (n = 6, p = 0.0883)

P [Z = 1] = 61*0.08831*0.91175= 0.3337

(d) The 10-minute intervals are consecutively numbered and observed whether overcrowded or otherwise. What is the probability that the 3rd overcrowded interval is the 5th interval, starting from 1 pm? (2 marks)

P [3rd overcrowded interval is the 5th interval, starting from 1 pm]
= P [there are 2 overcrowded intervals in the first four 10-minute intervals] * P [5th 10-minute interval is overcrowded]
= 42*0.08832*0.91172 * 0.0883 = 0.038858 * 0.0883 = 0.00343

(e) Given that on a particular day, the restaurant just barely avoids incurring a loss, what is the chance that no vehicle arrived in the first 10 minutes of the peak lunch hour on that day? (4 marks)

P [no arrivals in the 1st 10-minute interval | number of arrivals = 6 in one hour]
= P [no arrivals in the 1st 10-minute interval AND number of arrivals = 6 in one hour]/P [number of arrivals = 6 in one hour]
=P [no arrivals in the 1st 10-minute interval] *P [number of arrivals = 6 in next 50 minutes]/P [number of arrivals = 6 in one hour] = 0.334898

P [no arrivals in the 1st 10-minute interval] = P [Y = 0] = 0.188876
P [number of arrivals = 6 in one hour] = P [X = 6] = 0.063055458
Let W = number of arrivals in 50 minutes; W ~ Poisson (λ = 50/6)
P [number of arrivals = 6 in next 50 minutes] = P [W = 6] = 0.111805

2. The ABC company produces detergent packets of 1 kg each for the lower income segments. Its competitor announces that it guarantees that its 1 kg packets never fall short on weight. ABC knows that its filling machines are set at 1.10 kg and that the standard deviation of the quantity filled is 50 grams.

a. What percentage of the packets sold by ABC is below the 1 kg level?
(2 marks)

X = weight of detergent packet
X ~N (µ = 1.1, σ2 = 0.052)
P [X < 1] = 0.022750132

b. If ABC wants at most 1% to be below 1 kg, then where must it set the mean value for the filling machine? Assume that the standard deviation does not change.
(3 marks)
Find mu such that P [X < 1] ≤ 0.01

→ (1 – µ)/0.05 = -2.32634787 → µ = 1.116317394

c. The production manager comes up with an alternative. He says with an investment of Rs. 60,000, they can reduce the standard deviation to 35 grams and keep the mean setting at the same level of 1.10 kg. Assigning a cost of Rs.3 to each packet that is of lower weight than 1 kg (in terms of customer goodwill loss etc.), is this investment justified? Assume that they sell a million packets a year and would want to recover the investment in the first year of operations.
(4 marks)

P [X < 1| µ = 1.1, σ = 0.05] = 0.022750132
P [X < 1| µ = 1.1, σ = 0.035] = 0.002137
Saving in cost = Rs. 3 * 1000000 * (0.022750132 - 0.002137) = 61838.29

Investment is justified.
3. Let X be the amount (in thousand INR) of money that may be spent on advertisement for a product. It is decided that x, the possible values of X, can be: 0, 10 or 20. Let Y be the amount of additional profit (in lakh INR) that may be generated due to the advertisement. y, the possible values of Y can be: 0, 1, 2, or 3. The joint probability (density) function of these two discrete random variables X and Y is given by

a. Find the value of the constant c. Write down the joint distribution in a two-way table format. (2 marks)

| Y = 0 | Y = 100, 000 | Y = 200,000 | Y = 300, 000 | Marginal | X = 0 | 3c | 2c | c | 0 | 6c | X = 10, 000 | 5c | 4c | 3c | 2c | 14c | X = 20, 000 | 7c | 6c | 5c | 4c | 22c | Marginal | 15c | 12c | 9c | 6c | 42c |

42c = 1; so, c = 1/42

b. What is the probability that the amount of money spent on advertisement is more than or equal to INR 10,000 and the additional profit generated is less than or equal to INR 200,000? (1 mark)

P [X ≥ 10, 000, Y ≤ 200,000] = 30c /42c = 30/42= 5/7= 0.7143

c. What is the probability that the amount of money spent on advertisement is less than or equal to INR 10,000? (1 mark)

P [X ≤ 100,000] = 20c /42c = 20/42= 10/21= 0.4762

d. Given that the additional profit generated is INR 100,000 or more, what is the probability that the amount of money spent on advertisement is more than or equal to INR 10,000? (1.5 marks)

The conditional distribution of X, given Y ≥ 100,000, is:

| Y ≥ 100, 000 | X = 0 | 3/27 | X = 10, 000 | 9/27 | X = 20, 000 | 15/27 |

P [X ≥ 10, 000|Y ≥ 100,000] = 24/27 = 8/9 = 0.8889

e. Given that the amount of money spent on advertisement is INR 20,000, what is the expected value of the additional profit generated? (1.5 marks)

The conditional distribution of Y, given X = 20,000, is:

| Y = 0 | Y = 100, 000 | Y = 200,000 | Y = 300, 000 | X = 20, 000 | 7/22 | 6/22 | 5/22 | 4/22 |

E [Y|X = 10,000] = 0*7/22 + 100,000*6/22 + 200,000*5/22 + 300,000*4/22 = 127272.7 = 1.2727*105

f.
Compute the correlation coefficient and interpret it. (4 marks)

E[X] = 0*6/42 + 10,000*14/42 + 20,000*22/42 = 13809.524 = 1.381* 104

E[Y] = 114285.714 = 1.143*105

E[X 2] = 10,0002 * 14/42 + 20,0002 * 22/42 = 242857142.857 = 2.4286 * 108

Var (X) = E[X 2] – {E[X]}2 = 0.521439*108 SD(X) = 0.722107333*104

E[Y 2] = 100,0002 * 12/42 + 200,0002 * 9/42 +300,0002 * 6/42
= 24285714285.714 = 2.4286*1010

Var (Y) = E[Y 2] – {E[Y]}2 = 1.122151*1010
SD(Y) = 1.0593*105

E(XY) = 1.714286*109

Cov(X, Y) = 0.135803*109

Corr(X, Y) = 0.177534172

There is a weak positive correlation between X and Y.
4. A small farmer is trying to figure out whether he should use the high yielding variety (H) of seeds or the traditional (T) one. He has some idea that the costs will go up, and that the high yield will depend on the monsoon since his land is not irrigated. In addition to seed costs there are additional costs of fertilizer, and if he uses H, the total cost (of irrigation and seed, combined) per acre will be Rs.4,000, whereas if he uses T the total cost will be Rs.2000 per acre. The probability of excellent monsoon is 0.25, good monsoon has probability of 0.35, and bad monsoon has a probability of 0.4. If the monsoon is excellent, H gives a revenue of Rs.10,000, and T only Rs.4,000. If the monsoon is good, H gives a revenue of Rs.6,000, and T still gives Rs.4,000. If the monsoon is bad, H gives zero revenue, and T gives Rs.3,000. All the revenue figures are per acre and naturally do not take into account of cost of seed or irrigation.

a) What would you advise him to do in terms of variety of seeds to use? Give proper objective quantitative argument in support of your advice. (3 marks)

| Excellent | Good | Bad | | Prior Prob.→ | 0.25 | 0.35 | 0.4 | Expected Payoff | Use H | 6 | 2 | -4 | 0.6 | Use T | 2 | 2 | 1 | 1.6 |

Best Decision: Use T b)
Someone advises the Government that instead of giving subsidy, it should invest in better forecasting. He shows some data that indicates that if the monsoon is excellent, the forecast is also for excellent rain with probability 0.7, good rain with probability 0.2, and bad rain with probability 0.1. If the monsoon is good, the forecast is for excellent rain with probability 0.2, good rain with probability 0.6, and bad rain with probability 0.2. If the monsoon is bad, the forecast is for excellent rain with probability 0.05, good rain with probability 0.1, and bad rain with probability 0.85. How should the forecast information be used optimally? (Draw the complete decision tree with all necessary computations in support) What is the value of this information per acre for the farmer? (4+2 marks)

0.6604
0.2641
0.0755
0.6604
0.2641
0.0755
0.1667
0.7000
0.1333
0.1667
0.7000
0.1333
0.0575
0.1609
0.7816
0.0575
0.1609
0.7816
6
2
-4
6
2
-4
6
2
-4
2
2
1
2
2
1
2
2
1
4.1887
1.8667
-2.4598
1.2184
1.9245
1.8667
4.1887
1.8667
1.2184
0.265
0.435
X1
X2
X3
2.2
Forecast
D2
D1
D2
D2
D1
D1
 1
 2
 3
 1
 2
 3
 1
 2
 3
 1
 2
 3
 1
 2
 3
 1
 2
 3
Decisions
D1: Use H
D2: Use T
States of Nature
 1: Excellent Monsoon
 2: Good Monsoon
 3: Bad Monsoon
Outcomes of Forecast
X1: Forecast - Excellent
X2: Forecast - Good
X3: Forecast - Bad
P( i) P(X1 | i) P(X2 | i) P(X3 | i) 0.25 0.7 0.2 0.1
0.35 0.2 0.6 0.2
0.4 0.05 0.1 0.85

Payoffs in thousand Rs.
 1
 2
 3
LEGENDS
0.300

EVSI = 2.2 – 1.6 = 0.6

c) What is the value of perfect information? (2 marks)

EVPI = 0.25*6 + 0.35*2 + 0.4*1 – 1.6 = 1

5. A box has 40 items. Consider an experiment where two items are chosen at random, and if a particular designated item, say item A is selected, we stop. Else, we pick another pair of items from the remaining 38 items and continue like this till item A is picked.

a) What is the chance that we have to draw more than 3 times (i.e. more than 3 pairs)? (3 marks)
X = number of draws

PX=1=40-11402=40-124040-1=240=120
PX=2=1-12040-3140-22=20-12040-3240-240-3=120
PX=3=1-22040-5140-42=20-22040-5240-440-5=120

Continuing, we get

PX=i=120, for i=1, 2, …, 20

PX>3= 1-PX=1-PX=2-PX=3=1-320=1720=0.85

b) What is the expected number of draws before the item A is selected? (3 mark)

EX=1201+2+…+20=12020*212=10.5