Airplanes approaching the runway for landing are required to stay within the localizer (a certain distance left and right of the runway). When an airplane deviates from the localizer, it is sometimes referred to as an exceedence. Consider one airline at a small airport with six daily arrivals and an exceedence rate of 7%. a) Find the probability that on one day no planes have an exceedence. (2 points) b) Find the probability that at least 1 plane exceeds the localizer. (3 points) c) What is the expected number of planes to exceed the localizer on any given day? (2 points) d) What is the variance for the number of planes to exceed the localizer on any given day? (3 points)

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:

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.

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

12*c

step-by-step explanation: