Answer to Question #145948 in Statistics and Probability for Akhtiar

Question

Assume that a real estate manager has 5 contacts per week, and she believes that for each contact
the probability of making a sale is 0.30. Calculate the following if possible. If any part of question
(you feel) is not possible to calculate give reason:
a. Develop the probability distribution table for sale
b. Calculate the expected number of sales per week and variance of sale
c. Find the probability that she makes at most 2 sales.
d. Find the probability that she makes between 3 and 6 sales (both inclusive).

Answer 1

a. Let $X=$ the number of sales: $X\sim Bin(n, p)$

$$P(X=x)=\dbinom{n}{x}p^x(1-p)^{n-x}$$

Given $n=5, p=0.3$

$$P(X=0)=\dbinom{5}{0}0.3^0(1-0.3)^{5-0}=0.16807$$

$$P(X=1)=\dbinom{5}{1}0.3^1(1-0.3)^{5-1}=0.36015$$

$$P(X=2)=\dbinom{5}{2}0.3^2(1-0.3)^{5-2}=0.3087$$

$$P(X=3)=\dbinom{5}{3}0.3^3(1-0.3)^{5-3}=0.1323$$

$$P(X=4)=\dbinom{5}{4}0.3^4(1-0.3)^{5-4}=0.02835$$

$$P(X=5)=\dbinom{5}{5}0.3^5(1-0.3)^{5-5}=0.00243$$

$$\begin{matrix} x & p(x) \\ 0 & 0.16807 \\ 1 & 0.36015\\ 2 & 0.3087\\ 3 & 0.1323\\ 4 & 0.02835 \\ 5 & 0.00243 \end{matrix}$$

$P(X\geq6)=0$

2.

$$mean=E(X)=np=5(0.3)=1.5$$

$$Var(X)=np(1-p)=5(0.3)(1-0.3)=1.05$$

3.

$$P(X\leq2)=P(X=0)+P(X=1)+P(X=2)=$$

$$=0.16807+0.36015+0.3087=0.83692$$

4.

$$P(3\leq X \leq6)=$$

$$=P(X=3)+P(X=4)+P(X=4)+0=$$

$$=0.16308=P(X\geq3)$$