Question

An airline has found about 7% of its passengers request vegetarian meals. On a flight with 166 passengers the airline has 16 vegetarian dinners available. What's the probability that it will be short of vegetarian dinners? (hint: Let X be the number of vegetarians. Identify a probability model for X. Write a probability expression for X that reflects "Short of vegetarian dinners" and compute.)

Answer 1

P(x >16.5) = 0.3372

Step-by-step explanation:

Given data:

P = 0.07

n = 166

Available vegetarian dinner is 16

let
`P(X\leq 16)` is number of short vegetarian meals

`P(X\leq  16)` = binomial distribution (166, 0.09)

`np = 166* 0.09 = 14.94`

n(1-p) = 166(1-0.09) = 151.06

Both value of np and n(1-p) greater than 5

x - normal distribution with

mean = np = 14.94

standard deviation
`= √(np(1-p))`

[/tex]= \sqrt{14.94(1-0.09)}[/tex]

standard deviation = 3.687

Find P(x> 16) i.e P(X>16 ) = P(x >16.5)

P(x >16.5) = 1 - P(x <16.5)

`= 1 - P((x-\mu)/(\sigma) < (16.5 - \mu)/(\sigma)`

`= 1 - P{Z < [(16.5 - 14.94)/(3.67)]`

= 1 - P{z< 0.425}

= 1 - 0.6628

P(x >16.5) = 0.3372