Statistics Question!! Please please please please!!!!!!!!!! solve this problem w
ID: 3170512 • Letter: S
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.