a wedding planner does some research and finds that approximately 3.5% of the pe

Question

a wedding planner does some research and finds that approximately 3.5% of the people in the area where a large wedding is being held are pollotarian. Treat the 300 guests expected at the wedding as a simple random sample from the local population of about 200,000. Suppose the wedding planner assumes that only 5% of the guests will be pollotarian so she orders 15 pollotarian meals. What is the approximate probability that more than 5% of the guests are pollotarian and therefore not have enough meals?

a. 0.079

b. 0.421

c. 0.489

d. None of the above

Explanation / Answer

Answer : d. None of the above

Descriptive solution :

Binomial distribution with n = 300, p = 0.035
P(x) = 300Cx *0.035x *0.965(300-x)

Use formula & simplify computations using the complement,
P(>15) = 1 - [ P(0) +P(1) +..... +P(15) ] = 0.0648 <----

ps:
----
If by "approximate" you mean the normal approximation, i'll give you the method & the answers. you check !
n = 300, p = .035
(i) µ = np
(ii) = (npq)
(iii) with continuity correction, > 15 becomes > 15.5
(iv) z-score = (15.5 - µ)/ = k (say)
(v) look up a z-table to get P(z > k) = 0.0581

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