The response times of a customer service team for a particular company follow
a normal distribution with a mean of 30 minutes and a standard deviation of 6
minutes. What is the probability that at least one fifth of forty randomly selected
customers has to wait more than 35 minutes to receive a response? Use approximation with a normal approximation. 

To solve this problem, we need to use the normal approximation to the binomial distribution. Here are the steps to find the probability that at least one fifth of forty randomly selected customers has to wait more than 35 minutes to receive a response.

1. **Find the probability of one customer waiting more than 35 minutes:**

   The response times are normally distributed with a mean (\(\mu\)) of 30 minutes and a standard deviation (\(\sigma\)) of 6 minutes. We need to find the probability that a customer waits more than 35 minutes.

   \[
   Z = \frac{X - \mu}{\sigma} = \frac{35 - 30}{6} = \frac{5}{6} \approx 0.8333
   \]

   Using the standard normal distribution table, we find the probability corresponding to \(Z = 0.8333\). The cumulative probability for \(Z = 0.8333\) is approximately 0.797. Therefore, the probability of waiting more than 35 minutes is:

   \[
   P(X > 35) = 1 - 0.797 = 0.203
   \]

2. **Define the binomial problem:**

   We are dealing with a binomial distribution \(X \sim \text{Binomial}(n=40, p=0.203)\), where \(n\) is the number of trials (customers) and \(p\) is the probability of success (waiting more than 35 minutes).

3. **Normal approximation to the binomial distribution:**

   For large \(n\), a binomial distribution can be approximated by a normal distribution with mean \(\mu = np\) and standard deviation \(\sigma = \sqrt{np(1-p)}\).

   \[
   \mu = 40 \times 0.203 = 8.12
   \]
   \[
   \sigma = \sqrt{40 \times 0.203 \times (1 - 0.203)} \approx 2.578
   \]

4. **Find the z-score for at least one fifth of 40 customers:**

   One fifth of 40 customers is 8 customers. We need to find the probability that \(X \geq 8\).

   Applying continuity correction, we consider \(X \geq 7.5\).

   \[
   Z = \frac{7.5 - \mu}{\sigma} = \frac{7.5 - 8.12}{2.578} \approx -0.240
   \]

   Using the standard normal distribution table, we find the cumulative probability for \(Z = -0.240\).

   \[
   P(Z < -0.240) \approx 0.405
   \]

   Therefore, the probability that at least 8 out of 40 customers will have to wait more than 35 minutes is:

   \[
   P(X \geq 8) = 1 - 0.405 = 0.595
   \]

So, the probability that at least one fifth of forty randomly selected customers has to wait more than 35 minutes to receive a response is approximately \(0.595\).

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How do you find the z-score for a given value in a normal distribution?
2. What are the conditions for using the normal approximation to the binomial distribution?
3. How does continuity correction work in normal approximation?
4. How would the probability change if the standard deviation were different?
5. How does the central limit theorem apply to the approximation of binomial distribution by normal distribution?

**Tip:** Always check the conditions under which the normal approximation is valid, such as having a sufficiently large sample size and ensuring \(np\) and \(n(1-p)\) are both greater than 5.

The response times of a customer service team for a particular company follow a normal distribution with a mean of 30 minutes and a standard deviation of 6 minutes. What is the probability that at least one fifth of forty randomly selected customers has to wait more than 35 minutes to receive a response? Use approximation with a normal approximation.

Learn how to calculate the probability that at least one fifth of forty customers waits more than 35 minutes for a response, using normal approximation to binomial distribution. With a mean of 30 minutes and a standard deviation of 6, this problem covers concepts in probability, normal and binomial distributions, and continuity correction.

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