To solve this problem, we will use the binomial distribution to calculate the probability of at least covering the transportation costs and additional fee. The binomial distribution is useful because each customer either selects or does not select the brand, making it a series of independent Bernoulli trials.

Given data:

• n = 200 (number of customers)
• p = 0.28 (probability of a customer selecting the widget)
• X = the number of customers selecting the widget

We need to find the probability that at least 40 customers select the widget. In terms of the binomial distribution, we are looking for: $P(X \geq 40)$ where X follows a binomial distribution $X \sim \text{Binomial}(n=200, p=0.28)$.

Step 1: Calculate the parameters for the binomial distribution

The expected number of customers selecting the widget is: $\mu = np = 200 \times 0.28 = 56$ The standard deviation is: $\sigma = \sqrt{np(1-p)} = \sqrt{200 \times 0.28 \times (1 - 0.28)} = \sqrt{200 \times 0.28 \times 0.72} \approx 6.993$

Step 2: Use the normal approximation

For large n, the binomial distribution can be approximated by a normal distribution. Therefore, we approximate $X$ by a normal distribution: $X \sim \mathcal{N}(\mu = 56, \sigma \approx 6.993)$

We can apply the continuity correction by considering the probability that $X \geq 39.5$ (since $X \geq 40$ and we are using a continuous distribution).

Step 3: Standardize and calculate the z-score

We standardize the value $X = 39.5$ to find the z-score: $z = \frac{X - \mu}{\sigma} = \frac{39.5 - 56}{6.993} \approx \frac{-16.5}{6.993} \approx -2.36$

Step 4: Look up the z-score in the standard normal distribution table

Now, we find the cumulative probability for $z = -2.36$. From the standard normal distribution table, the cumulative probability for $z = -2.36$ is approximately 0.0091.

Step 5: Calculate the desired probability

The probability $P(X \geq 40)$ is the complement of $P(X \leq 39.5)$, which is: $P(X \geq 40) = 1 - P(X \leq 39.5) = 1 - 0.0091 = 0.9909$

Final Answer:

The probability that at least 40 customers select the widget is approximately 0.9909.

Would you like more details on any part of the solution, or do you have any other questions?

Here are a few follow-up questions:

1. How does the binomial distribution differ from the normal distribution in terms of application?
2. What is the continuity correction, and why is it important when approximating a binomial distribution?
3. How does the standard deviation of the binomial distribution affect the probability calculations?
4. What would happen if the probability of selection were higher or lower than 0.28?
5. How could you use this information to optimize widget sales in the store?

Tip: Always check if you can use the normal approximation to simplify calculations, especially for large sample sizes, like this case with 200 customers.