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

Answer:

P(x >16.5) = 0.3372

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)

$np = 166\times 0.09 = 14.94$

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$= \sqrt{np(1-p)}$

                                [/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)

                $= 1 - P(\frac{x-\mu}{\sigma} < \frac{16.5 - \mu}{\sigma}$

         $= 1 - P{Z < [\frac{16.5 - 14.94}{3.67}]$

                = 1 - P{z< 0.425}

                 = 1 - 0.6628

P(x >16.5) = 0.3372