In the following problem, check that it is appropriate to use the normal approxi

Question

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. What's your favorite ice cream flavor? For people who buy ice cream, the all-time favorite is still vanilla. About 23% of ice cream sales are vanilla. Chocolate accounts for only 11% of ice cream sales. Suppose that 166 customers go to a grocery store in Cheyenne, Wyoming, today to buy ice cream. (Round your answers to four decimal places.)

(a) What is the probability that 50 or more will buy vanilla?

(b) What is the probability that 12 or more will buy chocolate?

(c) A customer who buys ice cream is not limited to one container or one flavor. What is the probability that someone who is buying ice cream will buy chocolate or vanilla? Hint: Chocolate flavor and vanilla flavor are not mutually exclusive events. Assume that the choice to buy one flavor is independent of the choice to buy another flavor. Then use the multiplication rule for independent events, together with the addition rule for events that are not mutually exclusive, to compute the requested probability.

(d) What is the probability that between 50 and 60 customers will buy chocolate or vanilla ice cream? Hint: Use the probability of success computed in part (c).

Explanation / Answer

Solution:

(a) mean = n*p = 166*0.23 = 38.18

standard deviation = sqrt(166*0.23*(1-0.23)) = 5.4220

So the probability is

P(X > 50) = P((X-mean)/s >(50-38.18)/5.4220)

= P(Z > 2.18) = 0.0146

(b) mean = n*p = 166*0.11 = 18.26

standard deviation = sqrt(166*0.11*(1-0.11)) = 4.0313

So the probability is

P(X >12)= P(Z>(12-18.26)/4.0313) = P(Z> -1.55 ) = 0.9394

(c) The probability is

P(chocolate or vanilla) = P(chocolate) + P(vanilla) - P(both)

= 0.11+0.23-0.11*0.23 = 0.3147

(d) mean=166*0.3147 = 52.2402

standard deviation =sqrt(166*0.3147*(1-0.3147)) = 5.9833

So the probability is

P(50<X<60) = P((50-52.2402)/5.9833 <Z<(60-52.2402)/5.9833)

=P(-0.37 <Z< 1.29)

= 0.5458

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