Stats: Multiple Probabilities

• Dec 15th 2008, 06:59 PM
JeanetteR
Stats: Multiple Probabilities
2. Peter is a door-to-door sales representative who makes a sale at only 8% of his house calls. If Peter makes 219 house calls today, what is the probability that:
(a) he makes a sale at 12 or fewer houses?

Someone was trying to help me and said 12 x 100/219 = 5.6% but I have no idea where the 100/219 comes into play.

(b) he makes a sale at 20 or fewer houses?

(c) he makes a sale at 15 to 25 houses?
• Dec 15th 2008, 08:16 PM
mr fantastic
Quote:

Originally Posted by JeanetteR
2. Peter is a door-to-door sales representative who makes a sale at only 8% of his house calls. If Peter makes 219 house calls today, what is the probability that:
(a) he makes a sale at 12 or fewer houses?

Someone was trying to help me and said 12 x 100/219 = 5.6% but I have no idea where the 100/219 comes into play. Mr F says: This is wrong.

(b) he makes a sale at 20 or fewer houses?

(c) he makes a sale at 15 to 25 houses?

Let X be the random variable number of sales made.

X ~ Binomial(n = 219, p = 0.08)

(a) Calculate $\displaystyle \Pr(X \leq 12)$.

(b) Calculate $\displaystyle \Pr(X \leq 20)$.

A normal approximation to the binomial distribution would be valid here.
• Dec 15th 2008, 08:22 PM
tester85
For part C

You have to calculate

P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25)

X ~ Binomial(n = 219, p = 0.08)
to find P(x=15) = ( 219 C 15 )(0.08)^15(1-0.08)^204

Now you have to calculate the rest and add them together.