Question

1. a survey of 200 people attending a thanksgiving dinner asked about their favorite thanksgiving dish and whether they preferred white meat or dark meat turkey. the results are shown in the table.

what proportion of people who prefer white meat turkey also chose turkey as their favorite dish?

1. a local bakery sold a variety of thanksgiving pies. the distribution of the number of pies sold each day during november is approximately normal with a mean of 24 pies and a standard deviation of 5 pies.

what is the approximate number of pies sold on a day when the sales were at the 79th percentile?

1. a bakery offers three types of thanksgiving pies: pumpkin, apple, and pecan. 50% of their customers prefer pumpkin pie, 30% prefer apple pie, and 20% prefer pecan pie. if three customers are randomly selected, what is the probability that at least one of them prefers pecan pie?

Explanation:

Question 1

Step1: Identify relevant counts

People who prefer white meat turkey: 110 (from Total row, White Meat column). People who prefer white meat and chose turkey: 50 (from Turkey row, White Meat column).

Step2: Calculate proportion

Proportion = $\frac{\text{White meat + Turkey}}{\text{Total white meat}}$ = $\frac{50}{110} \approx 0.45$

Step1: Recall normal distribution percentile

79th percentile: Find z - score for 0.79. From z - table, z ≈ 0.81 (since P(Z < 0.81) ≈ 0.7910).

Step2: Use z - score formula

$z=\frac{x-\mu}{\sigma}$, so $x=\mu + z\sigma$. Given $\mu = 24$, $\sigma = 5$, $z = 0.81$. Then $x=24+(0.81\times5)=24 + 4.05 = 28.05\approx28$

Step1: Probability of not pecan

Probability a customer does not prefer pecan: $1 - 0.2=0.8$ (since P(pecan) = 0.2).

Step2: Probability none prefer pecan

For three independent customers, P(none prefer pecan) = $0.8\times0.8\times0.8 = 0.8^{3}=0.512$

Step3: Probability at least one pecan

P(at least one pecan) = $1 - P(\text{none})=1 - 0.512 = 0.488$

Answer:

C. 0.45

Question 2