Question

The arrival of flights ar DIA has been monitored for the last year. From the research, 65.17 % of all arrivals are on time. Suppose a random sample of 16 flight arrivals is examined.
Using the binomial function,answer the following questions.

1. Create a table and enter only the first and last value in that table.

k	P(X = k)
0
..	..
..	..
16

2. Give the probability of exactly 10 on time arrivals?


3. Give the probability of at most 9 on time arrivals?


4. Give the expected (mean) mean number of on time arrivals.

Answer 1

Here we have n = 16, p = 0.6517

We use excel to find probabilities . We use function "=binom.dist(x,n,p,0)" .

1.

k	p(X=k)
0	0.00000
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16	0.00106

2.

Here we need to find,

p ( x = 10 )

p ( X = x ) = ⁿC_x * p^x * ( 1 - p )^n-x

p ( X = 10 ) = ¹⁶C₁₀ * 0.6517¹⁰ * ( 1 - 0.6517 )^16-10

= 0.1976

3.

At most 9.

p ( x 9 ) = 1 - p ( x > 9 )

= 1 - { p ( x = 10) + p ( x =11 ) + p ( x =12 ) + p ( x = 13 ) + p ( x =14) + p ( x = 15) + p ( x = 16 ) }

= 1 - { ¹⁶C₁₀ * 0.6517¹⁰ * ( 1 - 0.6517 )^16-10 + ¹⁶C₁₁ * 0.6517¹¹ * ( 1 - 0.6517 )^16-11 + ¹⁶C₁₂ * 0.6517¹² * ( 1 - 0.6517 )^16-12 + ¹⁶C₁₃ * 0.6517¹³ * ( 1 - 0.6517 )^16-13 + ¹⁶C₁₄ * 0.6517¹⁴ * ( 1 - 0.6517 )^16-14 + ¹⁶C₁₅ * 0.6517¹⁵ * ( 1 - 0.6517 )^16-15 + ¹⁶C₁₆ * 0.6517¹⁶ * ( 1 - 0.6517 )^16-16 }

= 1 - { 0.1976 + 0.2016 + 0.1572 + 0.0905 + 0.0363 + 0.0091 + 0.0011 }

= 1 - 0.6934

= 0.3066

4.

Expected mean = np = 16 * 0.6517 = 10.43