Statistics Question!! Please please please please!!!!!!!!!! solve this problem w

Question

Statistics Question!!

Please please please please!!!!!!!!!! solve this problem with specific explanation and clear handwriting(or typing) please!

If you solve with your handwriting please solve with clear handwriting.....please....Sometimes I can't read you guys handwriting...

Thanks!

3. A salesman has scheduled two appointments to sell laptops. His first appointment will lead to a sale with probability .3, and his second will lead independently to a sale with proba- bility .6. Any sale made is equally likely to be either for the deluxe model, which costs $1000. or the standard model, which costs $500. Determine the probability mass function of X, the total dollar value of all sales.

Explanation / Answer

The range of X is {0,500,1000,1500,2000}.

If we use the following abbreviations for events:

F: the rst appointment leads to a sale;

S: the second appointment leads to a sale;

F1: getting a sale of 500 at the rst appointment;

F2: getting a sale of 1000 at the rst appointment;

S1: getting a sale of 500 at the second appointment;

S2: getting a sale of 1000 at the second appointment;

Then

P(Fc) = 1P(F) = 10.3 = 0.7,

P(Sc) = 1P(S) = 10.6 = 0.4,

Fi F, Si S (i = 1,2)

and

P(F1|F) = P(F2|F) = P(S1|S) = P(S2|S) = 0.5.

So for i = 1,2 we have

P(Fi) = P(Fi|F)P(F) + P(Fi|Fc)P(Fc) = 0.5×0.3 + 0 = 0.15, and

P(Si) = P(Si|S)P(S) + P(S1|Sc)P(Sc) = 0.5×0.6 + 0 = 0.3.

Since two appointments are independent of each other,

P(X = 0) = P(Fc and Sc)

= P(Fc)P(Sc)

= (1.3)(1.6)

= .28

P(X = 500) = P((F1 and Sc) or (Fc and S1))

= P(F1)P(Sc) + P(Fc)P(S1)

= 0.15 ×0.4 + 0.7×0.3

= .27

P(X = 1000) = P((F2 and Sc) or (F1 and S1) or (Fc and S2))

= P(F2)P(Sc) + P(F1)P(S1) + P(Fc)P(S2)

= 0.15×0.4 + 0.15×0.3 + 0.7×0.3

= .315;

P(X = 1500) = P((F1 and S2) or (F2 and S1))

= P(F1)P(S2) + P(F2)P(S1)

= 2×0.15×0.3

= 0.09

and

P(X = 2000) = P(F2 and S2)

= P(F2)P(S2)

= 0.15×0.3

= .045.

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