Prob.5 (a)Find the mean and variance of the random variable X if it is known tha

Question

Prob.5 (a)Find the mean and variance of the random variable X if it is known that EICX 1)1 10 and EI(X - 2)21 6 (b) A company purchases several special computers at the end of each year for a project, the exact number depending on the frequency of repairs in the previous year. Suppose that the number of computers, X, that are purchased each year has the following probability distribution 2 P (X = x) 1/10 3/10 2/5 1/5 If the cost of the desired model will remain fixed at $2500 throughout this year and a discount of 50X2 dollars is credited toward any purchase, how much can the company expect to spend on the computers at the end of this year?

Explanation / Answer

Q(a).> E[(X-1)^2]=E[X^2-2*X+1]=E[X^2]-2*E[X]+1 ----------(1)

E[(X-2)^2]=E[X^2-4*X+4]=E[X^2]-4*E[X]+4 -----------(2)

Equation (1)=10 and Equation (2)=6. -----------(3)

Now assume, u=E[X^2] and v=E[X]. ------------(4)

Using (1), (2), (3) and (4),

E[X^2]-2*E[X]+1=10 , => u-2*v+1=10

E[X^2]-4*E[X]+4=6, => u-4*v+4=6

Therefore, u=16 and v=7/2, => v=E[X]=7/2 and u=E[X^2]=16.

Mean=E[X]=v=7/2=3.5

Variance=E[(X-E[X])^2]=E[(X-7/2)^2]=E[X^2-7*X+49/4]=E[X^2]-7*E[X]+49/4=16-7*7/2+49/4=3.75

Therefore mean=3.5 and variance=3.75

Q (b). > Total payable for any purchase=2500-50*X^2, for a computer, 50*X^2 as discount for a computer. Now for x computers, we have X*(2500-50*X^2). Therefore,

One needs to find E[X*(2500-50*X^2)] for Xi=[0, 1, 2, 3]

Formula, E[X]=SUM(Xi*p(X=x))

Using above formula, E[X*(2500-50*X^2)]=SUM ( X*(2500-50*X^2)*p(X) ) for X[0, 1, 2, 3]

=0+1*(2500-50*1^2)*3/10+2*(2500-50*2^2)*2/5+3*(2500-50*3^2)*1/5

=3805

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