Answer:
a) P(x = 0) = 64.69%
b) P(x ≥ 1) = 35.31%
c) E(x) = 0.42
d) var(x) = 0.3906
Step-by-step explanation:
The given problem can be solved using binomial distribution since:
- There are n repeated trials independent of each other.
- There are only two possibilities: exceedence happens or exceedence doesn't happen.
- The probability of success does not change with trial to trial.
The binomial distribution is given by
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.
For the given scenario. the six daily arrivals are the number of trials
Number of trials = n = 6
The probability of success = 7% = 0.07
a) Find the probability that on one day no planes have an exceedence.
Here we have x = 0, n = 6 and p = 0.07
P(x = 0) = ⁶C₀(0.07⁰)(1 - 0.07)⁶⁻⁰
P(x = 0) = (1)(0.07⁰)(0.93)⁶
P(x = 0) = 0.6469
P(x = 0) = 64.69%
b) Find the probability that at least 1 plane exceeds the localizer.
The probability that at least 1 plane exceeds the localizer is given by
P(x ≥ 1) = 1 - P(x < 1)
But we know that P(x < 1) = P(x = 0) so,
P(x ≥ 1) = 1 - P(x = 0)
We have already calculated P(x = 0) in part (a)
P(x ≥ 1) = 1 - 0.6469
P(x ≥ 1) = 0.3531
P(x ≥ 1) = 35.31%
c) What is the expected number of planes to exceed the localizer on any given day?
The expected number of planes to exceed the localizer is given by
E(x) = n×p
Where n is the number of trials and p is the probability of success
E(x) = 6×0.07
E(x) = 0.42
Therefore, the expected number of planes to exceed the localizer on any given day is 0.42
d) What is the variance for the number of planes to exceed the localizer on any given day?
The variance for the number of planes to exceed the localizer is given by
var(x) = n×p×q
Where n is the number of trials and p is the probability of success and q is the probability of failure.
var(x) = 6×0.07×(1 - 0.07)
var(x) = 6×0.07×(0.93)
var(x) = 0.3906
Therefore, the variance for the number of planes to exceed the localizer on any given day is 0.3906.
