To determine if we can use the normal approximation to the binomial distribution in this problem, let's review the criteria and then apply the approximation.

Step 1: Verify Normal Approximation Criteria

The normal approximation to the binomial is appropriate when:

1. The sample size $n$ is large.
2. Both $np$ and $n(1 - p)$ are greater than or equal to 5.

Let's define the parameters:

• Number of trials (customers) $n = 173$
• Probability of success (buying vanilla) $p = 0.25$

Let's calculate $np$ and $n(1 - p)$ to check the conditions:

• $np = 173 \times 0.25 = 43.25$
• $n(1 - p) = 173 \times (1 - 0.25) = 129.75$

Both $np$ and $n(1 - p)$ are greater than 5, so it is appropriate to use the normal approximation.

Step 2: Apply the Normal Approximation

For a binomial distribution $B(n, p)$, we approximate it with a normal distribution $N(\mu, \sigma^2)$ where:

• $\mu = np$
• $\sigma = \sqrt{np(1 - p)}$

Calculate Mean and Standard Deviation

• $\mu = 43.25$
• $\sigma = \sqrt{43.25 \times 0.75} = \sqrt{32.4375} \approx 5.6963$

So, we use the normal distribution $N(43.25, 5.6963^2)$.

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If you provide the specific probability ranges you'd like to find (e.g., "more than 40 customers" or "between 30 and 50 customers"), I can calculate the probabilities accordingly.

Would you like further details or have any specific questions?

Here are 5 related questions for further exploration:

1. What is the probability that fewer than 30 customers will buy vanilla ice cream?
2. What is the probability that more than 50 customers will buy vanilla ice cream?
3. How does changing the probability of vanilla purchases affect the distribution's mean and standard deviation?
4. Can we apply the normal approximation to chocolate sales as well?
5. How would we approach this problem if only 50 customers visited the store instead of 173?

Tip: When using a normal approximation for a binomial, consider applying a continuity correction by adjusting the interval by 0.5, which can improve accuracy.