91Ó°ÊÓ

We have the probability of success (preferring Pepsi) as \(p = 0.25\). The number of consumers in a group is \(n=500\). The mean (\(\mu\)) and standard deviation (\(\sigma\)) for the binomial distribution are given by: Mean: \(\mu = np\) Standard Deviation: \(\sigma = \sqrt{np(1-p)}\) Let's plug in the values and calculate the mean and standard deviation: Mean: \(\mu = 500 \times 0.25 = 125\) Standard Deviation: \(\sigma = \sqrt{500 \times 0.25 \times 0.75} = 9.682\)

To solve each part, we need to calculate the Z-scores and use a standard normal table or a calculator to find the corresponding probabilities. The Z-score formula is given by: \(Z = \frac{X - \mu}{\sigma}\), where X is the number of consumers who prefer Pepsi. a. Exactly 150 consumers prefer Pepsi: \(Z = \frac{150 - 125}{9.682} = 2.582\) Since "exactly" is not a range and the normal distribution is continuous, we need to use the continuity correction: P(149.5 < X < 150.5). To do this, we calculate the Z-scores for 149.5 and 150.5 respectively as follows: \(Z_{149.5} = \frac{149.5 - 125}{9.682} = 2.528\) \(Z_{150.5} = \frac{150.5 - 125}{9.682} = 2.634\) Using a standard normal table or calculator, we find P(2.528 < Z < 2.634) ≈ 0.0045. b. Between 120 and 150 consumers (inclusive) prefer Pepsi: We will again use the continuity correction and calculate the Z-scores for 119.5 and 150.5: \(Z_{119.5} = \frac{119.5 - 125}{9.682} = -0.567\) \(Z_{150.5} = \frac{150.5 - 125}{9.682} = 2.634\) Now, we find P(-0.567 < Z < 2.634) ≈ 0.7165. c. Fewer than 150 consumers prefer Pepsi: For this part, we will use the continuity correction and calculate the Z-score for 149.5: \(Z_{149.5} = \frac{149.5 - 125}{9.682} = 2.528\) Now, we find P(Z < 2.528) ≈ 0.9942. d. To find whether it would be unusual to have 232 out of 500 consumers prefer Pepsi, we will calculate the Z-score for 232: \(Z = \frac{232 - 125}{9.682} = 11.03\) Now, let's consider the empirical rule, which states that 99.7% of the data falls within 3 standard deviations from the mean. Since 11.03 is way beyond 3 standard deviations, we can conclude that it would be extremely unusual to find that 232 of the 500 consumers preferred a Pepsi product. If this were to occur, one might question the initial assumption that the preference for Pepsi products is 25% of the market share, or suspect that the sample was biased.