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MATH1715 Tutee Revision Questions

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Question: 29/12/2010<br />

I am not quite sure how to obtain the answer for the following question.<br />

The demand for a certain weekly magazine at a news stand is a random variable with<br />

probability mass function pX(x) given by<br />

x 3 4 5 6<br />

pX(x) 0.20 0.44 0.32 0.04<br />

The magazine sells for £1.50, and the cost to the owner is 50 pence per copy. Unsold copies<br />

cannot be returned to the publisher.<br />

(i) What is the expected weekly demand for this magazine?<br />

(ii) If the owner orders three copies of the magazine, his net profit will be £3.00. What will<br />

be his expected profit if he orders four copies?<br />

(iii) How many copies should be ordered per week in order to maximise the expected profit?<br />

Response: Here X denotes the number of copies demanded per week.<br />

(i) E[X] = �<br />

xpX(x) = µ say.<br />

x<br />

(ii) Let N denote the number of copies he orders. To see how to tackle the question, look<br />

at some specific cases.<br />

If N = 3, then he pays 3 × 50p = £1.50. He sells them all if X ≥ 3, which occurs with<br />

probability unity. He receives 3 × 1.50 = £4.50, so his net profit is £3 as specified.<br />

If N = 4, then he pays 4 ×50p = £2.00. There are two cases to now consider. If X = 3,<br />

with probability 0.2, he sells 3, making net profit P3. If X ≥ 4, with probability 0.8, he sells<br />

all 4, making net profit P4. His net profit P then satisfies E[P] = (0.2 × P3) + (0.8 × P4).<br />

(iii) Repeat the calculations for N = 5 and N = 6. For example, for N = 5 you can<br />

determine what he has spent buying the copies. For each of the cases X = 3, X = 4, X ≥ 5<br />

you can determine his net profit and the associated probability. Thus you can determine<br />

the mean profit if N = 5. Similarly for N = 6.<br />

You could also consider the cases N = 0, 1, 2 and N = 7, 8, 9, . . .. In the latter case it<br />

is easy to see that his costs increase while his return is capped by the maximum demand<br />

X = 6; thus his net profit must certainly be monotonic decreasing for all N > 6. For N = 0<br />

his net profit is zero (a safe risk-free place to be!). For 0 ≤ N ≤ 3 it can be seen that his<br />

profit must be monotonic increasing as he always sells at least three copies. The maximum<br />

must therefore lie between N = 3 and N = 6.<br />

Question: 26/12/2010<br />

Is it a general rule that Var[aX + bY ] = a Var[X] + b 2 Var[Y ] + 2ab cov(X, Y )?<br />

Response: Yes.<br />

Question: 26/12/2010<br />

We toss a coin repeatedly until we see each side of the coin at least once. What is the mean<br />

number of tosses required?<br />

Response: Let X denote the number of tosses until we see each side at least once. What<br />

is the distribution of X? What is E[X]?<br />

10

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