Question

Two of the biggest soft drink rivals, Pepsi and Coke, are very concerned about their market shares. The pie chart that follows claims that PepsiCo's share of the beverage market is \(25 \% .^{6}\) Assume that this proportion will be close to the probability that a person selected at random indicates a preference for a Pepsi product when choosing a soft drink. A group of \(n=500\) consumers is selected and the number preferring a Pepsi product is recorded. Use the normal curve to approximate the following binomial probabilities. a. Exactly 150 consumers prefer a Pepsi product. b. Between 120 and 150 consumers (inclusive) prefer a Pepsi product. c. Fewer than 150 consumers prefer a Pepsi product. d. Would it be unusual to find that 232 of the 500 consumers preferred a Pepsi product? If this were to occur, what conclusions would you draw?

Short answer

In summary, using the normal distribution to approximate the binomial probabilities, we found that the probability of exactly 150 consumers preferring Pepsi is approximately 0.0045, the probability of between 120 and 150 consumers preferring Pepsi is approximately 0.7165, and the probability that fewer than 150 consumers prefer Pepsi is approximately 0.9942. Furthermore, having 232 out of 500 consumers prefer Pepsi would be extremely unusual, as it is far beyond three standard deviations from the mean, pointing to a potential issue with the initial assumption or sample bias.

Step-by-step solution

Calculate the Mean and Standard Deviation

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: \(\sigma = \sqrt{500 \times 0.25 \times 0.75} = 9.682\)

Calculate the Z-Scores and Corresponding Probabilities for Each Part

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.