The owners of a motel in Florida have noticed that in the long run, about 40% of the people who stop and inquire about a room for the night actually rent a room. (a) How many inquires must the owner answer to be 99% sure of renting at least one room? inquiries (b) If 20 separate inquiries are made about rooms, what is the expected number of inquiries that will result in room rentals? inquiries

(a) n = 9 inquiries

The owner must answer 9 inquiries to be 99% sure of renting at least one room.

(b) E(x) = 8 inquiries

Therefore, the expected number of inquiries that will result in room rentals is 8.

Step-by-step explanation:

The given problem can be solved by Binomial distribution.

Let x be the people who stop and inquire about a room for the night actually rent a room.

P(x) = ⁿCₓ pˣ (1 - p)ⁿ⁻ˣ  

Where n is the number of trials, x is the variable of interest and p is the probability of success.  

(a) How many inquires must the owner answer to be 99% sure of renting at least one room?

For the given case, we have p = 0.40 and we want to find n such that

P(x ≥ 1) = 0.99

we know that

P(x ≥ 1) = 1 - P(x < 1)

P(x ≥ 1) = 1 - P(x = 0)

So,

1 - P(x = 0) = 0.99

P(x = 0) = 1 - 0.99

P(x = 0) = 0.01

For x = 0 and p = 0.40

P(x = 0) =(ⁿC₀) (0.40)⁰ (1 - 0.40)ⁿ⁻⁰

0.01 = (1)(1)(0.60)ⁿ

0.01 = 0.60ⁿ

ln(0.01) = ln(0.60ⁿ)

ln(0.01) = n ln(0.60)

n = ln(0.01)/ln(0.60)

n = 9.01

Rounding off to nearest whole number

n = 9 inquiries

Therefore, the owner must answer 9 inquiries to be 99% sure of renting at least one room.

(b) If 20 separate inquiries are made about rooms, what is the expected number of inquiries that will result in room rentals?

The expected number of inquiries is given by

E(x) = n×p

For the given case, we have n = 20 and p = 0.40

E(x) = 20×0.40

E(x) = 8 inquiries

Therefore, the expected number of inquiries that will result in room rentals is 8.

idk lol

step-by-step explanation: