Exercise 1

Five percent  (5%)  of Christmas tree  light bulbs manufactured by  a  company  are defective. The company's Quality Control Manager is quite concerned and therefore randomly samples 100 bulbs coming off of the assembly line. Let X denote the number in the sample that are defective. Calculate the probability that the sample contains at most three defective bulbs. In doing this, determine which probability distribution should be used and eventually compare your result to that from an approximating distribution.

Answer:

There are two possible outcomes (defective or not), the 100 trials of selecting the bulbs from the assembly line can be assumed to be performed in an identical and independent manner, and the probability of getting a defective bulb can be assumed to be constant

from trial to trial. So, Xis a binomial random variable.

Just need to use the binomial with n = 100 and p = 0.05.

100                                     100

0                                         3

= 25.78%

or equivalently P(X ≤ 3) = 0.0059 + 0.0312 + 0.0811 + 0.1396 = 0.2578.

Since here the number of trials is high and the probability of success is very small, such that  入 = pn = 0.05 ∗ 100 = 5 < 7,  the  Poisson  distribution  can  also  be  used  to approximate the Binomial. Thus let's calculate P(X ≤ 3) by using the Poisson and see how close we get.

P(X ≤ 3) = e −5 ( +  +  + ) = 0.265

When we used the binomial distribution, we deemed P(X ≤ 3) ≌ 0.258, and when we used  the  Poisson  distribution,  we  deemed P(X ≤  3) =  0.265.  Not  too  bad  of an approximation!

Exercise 2

In the city of Beijing, the car accidents occur randomly at an average rate of 14 accidents per week. Then answer the following two questions:

a)  What is the probability of observing 8 accidents in a given week in Beijing?

b)  What is the probability of observing at least 4 accidents in a given day?

Answer:

Here  we  have  to  use  the  Poisson  distribution,  which  is  suitable  to  describe  the distribution of rare events in a large population.

a)   Car  accidents  occur  randomly  at  an  average  rate  of  14  accidents  per  week. Therefore, P(X = 8) = e −14 ≅ 0.0304.

b)  Car accidents occur randomly at an average rate of 2 accidents per day, therefore:

P(X ≥ 4) = 1 − P(X ≤ 3)

= 1 − [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

= 1 − [e −2 ∙ ( +  +  + )] ≅ 0.1428 ≅ 14.3%.

Exercise 3

A car salesperson estimates the following probabilities for the number of cars that sold each week.

Answers:

(a) Find the expected number of cars sold per week.

E(X) = p(X)·X

E(X) = 2 cars

(b) Find the standard deviation of the number of cars sold per week.

Var(X) = p(X)∙ [X – E(X)]2 = 1.78

SD(X) = √Va(X) = √1.78 = 1.334 cars

(c) The salesperson receives a salary of £250 each week, plus an additional £300 per car sold.  Find the mean and standard deviation of her weekly salary.

E(250 + 300X) = 250 + 300 ·E(X) = £ (250 + 600)

E(salary) = £850

SD (250 + 300 X) = 300 SD(X) = £ 300 ∙ 1,334

SD(salary) = £400.2

Exercise 4

A food company is considering using the media for advertising its products. Let X denote the number of media. Let Y denote the company’s number of product types. Using historical records, the joint probability distribution is shown in the following table.

You are required to

a)    Find the conditional probability distribution for Y, given X=3.

b)    Find the joint cumulative probability for (X=3, Y=2).

Answer:

a)

PY|X=3(Y = 1) =                                                     =

= 0.01 + 0.04 + 0.05 = 0.10 = 0.1 = 10%

PY|X=3(Y = 2) =                                                     =

= 0.01 + 0.04 + 0.05 = 0.10 = 0.40 = 40%

PY|X=3(Y = 3) =                                                     =

= 0.01 + 0.04 + 0.05 = 0.10 = 0.50 = 50%

b)   Joint cumulative probability at X=3, Y=2:

F(3,2) = P(X<=3,Y<=2) = P(X=1,Y=1) + P(X=1,Y=2) + P(X=2,Y=1) +

P(X=2,Y=2) + P(X=3,Y=1) + P(X=3,Y=2) = 0. 12+0. 19+0. 17+0.20+0.01+0.04 = 0.73 = 73%

Based on the data above, find out:

i.  The expected value of XY;

ii.  The expected value of X and the expected value of Y;

iii.  The covariance between X and Y.

iv.  Are number of media and number of product types statistically independent of one another? Why?

Answer:

i.      E(XY)=1*0. 12+2*0. 19+3*0. 10+2*0. 17+4*0.20+6*0. 12+3*0.01+6*0. 04+ 9*0.05 = 3.38;

ii.      E(X)=1*(0. 12+0. 19+0. 10)+2*(0. 17+0.20+0. 12)+3*(0.01+0.04+0.05)= 1.69;                                                                                                         E(Y)=1*(0. 12+0. 17+0.01)+2*(0. 19+0.20+0.04)+3*(0. 10+0. 12+0.05)= 1.97;

iii.      Cov(X,Y)=E(XY)-E(X)*E(Y)=3.38- 1.69*1.97=0.0507.

iv.      They are not independent as Cov(X,Y)≠0, although the level of dependence is quite low.

Exercise 5

An investor puts $30,000 into each of four stocks, labelled A, B, C, and D. The table shown below contains the means and standard deviations of the annual returns of these four stocks.

Assuming that the returns of the four stocks are not independent of one another, with the correlations between all pairs of stock returns being given in table below, find the mean and standard deviation of the total amount that this investor earns in one year from these four investments.

Answer:   Let Rj   be the annual return on stockj.

Total return = $30,000 ( RA  + RB  + RC  + RD )

Therefore the mean is given by:

E(Total Return) = $30,000 [ E(RA ) + E(RB ) + E(RC ) + E(RD ) ] = $30,000 [0.20 + 0.10 + 0.12 + 0. 16] = $17400.

For the  second  step, we  firstly  calculate the  covariances between  stock returns (variances  of stock returns  are  on the  diagonal) by using  formula  C0V(AB) = pA,B A B :

Stock A    Stock B   Stock C   Stock D

REMEMBER:

Varp,x      =    xA(2) ∙ uA(2) + xB(2) ∙ uB(2) + x C(2) ∙ uC(2) + xD(2) ∙ uD(2) + 2 ∙ [xAxB ∙ C0V(A, B) + xAxC ∙ C0V(A, C) + xAxD ∙ C0V(A, D) + xBxC ∙ C0V(B, C) + xBxD ∙ C0V(B, D) + xCxD ∙ C0V(C, D)] = 300002 ∙ uA(2) + 300002 ∙ uB(2) + 300002 ∙ uC(2) + 300002 ∙ uD(2) + 2 ∙      [300002 ∙ C0V(A, B) + 300002 ∙ C0V(A, C) + 300002 ∙ C0V(A, D) + 300002 ∙   C0V(B, C) + 300002 ∙ C0V(B, D) + 300002 ∙ C0V(C, D)].

for a given vector X = (xA, xB, xC, xD) of weights (investments) with xA = xB = xC = xD = 30,000. To simplify calculations, we build a matrix of variances and pairwise covariances related to the single investments:

30,000

30,000

30,000

30,000

30,000     30,000     30,000     30,000