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.)

P(x >16.5) = 0.3372

Explanation:

Given data:

P = 0.07

n = 166

Available vegetarian  dinner is 16

let is number of short vegetarian meals

 = binomial distribution (166, 0.09)

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

                                [/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{z< 0.425}

                 = 1 - 0.6628

P(x >16.5) = 0.3372

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