An airline has found that on average, 12% of its passengersrequest vegetarian me

Question

An airline has found that on average, 12% of its passengersrequest vegetarian meals. On a flight with 340 passengers theairline has 45 vegetarian dinners available. What's the probabilitythat it will be short of vegetarian dinners? Find the indicated probability by using an appropriate normalmodel to approximate the binomial distribution. A) 0.067 B) 0.919 C) 0.192 D) 0.081 E) 0.394 An airline has found that on average, 12% of its passengersrequest vegetarian meals. On a flight with 340 passengers theairline has 45 vegetarian dinners available. What's the probabilitythat it will be short of vegetarian dinners? Find the indicated probability by using an appropriate normalmodel to approximate the binomial distribution. A) 0.067 B) 0.919 C) 0.192 D) 0.081 E) 0.394

Explanation / Answer

Since the count is very large, if we used binomialprobability, we would have to calculate the probability of 0successes, 1, 2, 3, 4, 5, 6....all the way to 45, we will use thenormal approximation.
The standard deviation and mean of x can be determined withthe information given.
mean = number of observations * probability of success, whichis .12 * 340 = 40.8
standard deviation = sqrt[number of observations * probabilityof success *(1 - probability of success)] = sqrt(340 * .12 * .88) =5.99199.
We can use a z-table to approximate the normal (it might beslightly off).
So the z-score is (45-40.8)/5.99199 = .700936.
Looking this up in the table, we have 75.80% of values fallbelow that, so we have 24.20% fall above 45.
That is close to 0.192 - so I would pick that as theanswer.

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