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%. 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. (Round your answers to four decimal places.)

(a) Exactly 150 consumers prefer a Pepsi product.

(b) Between 110 and 150 consumers (inclusive) prefer a Pepsi product.

(c) Fewer than 150 consumers prefer a Pepsi product.

ANSWERS:
a) 0.0015
b) 0.9411
c) 0.9943

How do I get to these results? Thanks! :)

Never mind, figured it out :)

To find the binomial probabilities using the normal curve, we can use the normal approximation to the binomial distribution. The mean (μ) of the binomial distribution is given by:

μ = n * p

where n is the number of trials or consumers in this case, and p is the probability of success (proportion of consumers who prefer a Pepsi product).

The standard deviation (σ) of the binomial distribution is given by:

σ = sqrt(n * p * (1 - p))

Now, let's calculate the values for the given scenario:

n = 500 (the number of consumers)
p = 0.25 (PepsiCo's share of the beverage market, or the probability of a person preferring a Pepsi product)

(a) Exactly 150 consumers prefer a Pepsi product:
We need to calculate P(X = 150), where X is the random variable representing the number of consumers who prefer a Pepsi product. Using the normal approximation, we will calculate P(149.5 < X < 150.5), as we cannot have a fraction of a consumer.

To calculate this probability, we need to standardize the values using the formula:
Z = (X - μ) / σ

Now, let's calculate Z:
Z = (150 - (500 * 0.25)) / sqrt(500 * 0.25 * (1 - 0.25))

Next, we can find the probability using a standard normal distribution table, where we look up the probability associated with the calculated Z value. Since we need the probability between 149.5 and 150.5, we will subtract the probability of Z less than 149.5 from the probability of Z less than 150.5.

(b) Between 110 and 150 consumers (inclusive) prefer a Pepsi product:
We need to calculate P(109.5 < X < 150.5). First, we will standardize the lower and upper values using the Z formula.

Lower Z = (110 - (500 * 0.25)) / sqrt(500 * 0.25 * (1 - 0.25))
Upper Z = (150 - (500 * 0.25)) / sqrt(500 * 0.25 * (1 - 0.25))

To find the probability, we will subtract the probability of Z less than 109.5 from the probability of Z less than 150.5.

(c) Fewer than 150 consumers prefer a Pepsi product:
We need to find P(X < 150). Again, we will standardize the value using the Z formula.

Z = (150 - (500 * 0.25)) / sqrt(500 * 0.25 * (1 - 0.25))

We can then find the probability of Z less than 150.

Using these steps, you can calculate the probabilities and compare them with the provided answers.

To solve these problems, you can use the normal distribution to approximate the binomial probabilities. This approximation is valid when the sample size is large (n ≥ 30) and the success probability (p) is not too close to 0 or 1.

To find these probabilities, we will need to calculate the mean (μ) and the standard deviation (σ) of the binomial distribution. For a binomial distribution, the mean μ is equal to n * p, and the standard deviation σ is equal to √(n * p * (1 - p)).

Given that the proportion of PepsiCo's share of the beverage market is 25% (or 0.25) and the number of consumers selected is n = 500, we can calculate the mean and standard deviation as follows:

Mean (μ) = n * p = 500 * 0.25 = 125

Standard Deviation (σ) = √(n * p * (1 - p)) = √(500 * 0.25 * 0.75) ≈ 8.6603

(a) To find the probability that exactly 150 consumers prefer a Pepsi product, we are looking for P(X = 150), where X follows a binomial distribution with mean μ = 125 and standard deviation σ ≈ 8.6603.
We can use the normal distribution to approximate this probability by converting the discrete value of 150 to a continuous value using the continuity correction.

P(X = 150) ≈ P(149.5 < X < 150.5)

To standardize this range, we use the z-score formula: z = (X - μ) / σ

z1 = (149.5 - 125) / 8.6603 ≈ 2.8284
z2 = (150.5 - 125) / 8.6603 ≈ 2.9142

Using a standard normal distribution table or calculator, find the probabilities corresponding to the z-scores z1 and z2 and subtract them: P(149.5 < X < 150.5) ≈ P(z2) - P(z1) ≈ 0.9985 - 0.9969 ≈ 0.0015

Therefore, the probability that exactly 150 consumers prefer a Pepsi product is approximately 0.0015.

(b) To find the probability that between 110 and 150 consumers (inclusive) prefer a Pepsi product, we are looking for P(110 ≤ X ≤ 150), where X follows a binomial distribution with mean μ = 125 and standard deviation σ ≈ 8.6603.

Similarly, we convert the discrete values of 110 and 150 to continuous values using the continuity correction:

P(110 ≤ X ≤ 150) ≈ P(109.5 < X < 150.5)

To standardize this range, we use the z-score formula:

z1 = (109.5 - 125) / 8.6603 ≈ -1.7944
z2 = (150.5 - 125) / 8.6603 ≈ 2.9142

Using a standard normal distribution table or calculator, find the probabilities corresponding to the z-scores z1 and z2 and subtract them: P(109.5 < X < 150.5) ≈ P(z2) - P(z1) ≈ 0.9985 - 0.0359 ≈ 0.9411

Therefore, the probability that between 110 and 150 consumers (inclusive) prefer a Pepsi product is approximately 0.9411.

(c) To find the probability that fewer than 150 consumers prefer a Pepsi product, we are looking for P(X < 150), where X follows a binomial distribution with mean μ = 125 and standard deviation σ ≈ 8.6603.

To standardize this value, we use the z-score formula:

z = (149.5 - 125) / 8.6603 ≈ 2.8284

Using a standard normal distribution table or calculator, find the probability corresponding to the z-score z: P(X < 150) ≈ P(z) ≈ 0.9985

Therefore, the probability that fewer than 150 consumers prefer a Pepsi product is approximately 0.9985.

To summarize:

(a) P(X = 150) ≈ 0.0015
(b) P(110 ≤ X ≤ 150) ≈ 0.9411
(c) P(X < 150) ≈ 0.9985