To determine if it's appropriate to use the normal approximation to the binomial distribution, we should verify that the conditions for this approximation are met. The normal approximation is suitable when:

1. $np \geq 5$
2. $n(1 - p) \geq 5$

where:

• $n = 173$ (total customers),
• $p = 0.25$ (probability a customer buys vanilla).

Step 1: Check Conditions

1. Calculate $np$: $np = 173 \times 0.25 = 43.25$
2. Calculate $n(1 - p)$: $n(1 - p) = 173 \times (1 - 0.25) = 173 \times 0.75 = 129.75$

Both values are greater than 5, so we can proceed with the normal approximation.

Step 2: Define the Normal Approximation

For a binomial distribution $X \sim B(n, p)$, the normal approximation is $X \approx N(\mu, \sigma^2)$, where:

• $\mu = np = 43.25$,
• $\sigma = \sqrt{np(1 - p)} = \sqrt{43.25 \times 0.75} \approx 5.6944$.

Step 3: Find the Probability for $X \geq 50$

Using the normal approximation, we apply a continuity correction: $P(X \geq 50) \approx P\left(Z \geq \frac{50 - 0.5 - \mu}{\sigma}\right)$ where $Z$ is a standard normal variable.

Calculations

1. Calculate the z-score: $Z = \frac{49.5 - 43.25}{5.6944} \approx 1.0966$
2. Find $P(Z \geq 1.0966)$: Using a standard normal distribution table or calculator, $P(Z \geq 1.0966) \approx 0.1369$.

Answer

The probability that 50 or more customers will buy vanilla is approximately 0.1369.

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Do you want more details on any step, or have further questions?

Here are related questions that could extend this problem:

1. What is the probability that fewer than 50 customers buy vanilla?
2. What is the probability that exactly 43 customers buy vanilla?
3. How would the probability change if $p$ was increased to 0.3?
4. If 200 customers entered the store, what would the new probability for 50 or more buying vanilla be?
5. What is the mean and variance of the binomial distribution in this scenario?

Tip: When using the normal approximation to the binomial, remember to apply the continuity correction for more accurate results.