Respuesta :
Answer:
(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.
