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We are given that the number of loan applications is normally distributed with:

- Mean \(\mu = 66.4\)
- Standard deviation \(\sigma = 10.9\)

We need to find the following probabilities:

**a. More than 75 applications?**

To find this, we need to calculate the Z-score for 75 applications:

\[
Z = \frac{X - \mu}{\sigma} = \frac{75 - 66.4}{10.9} = \frac{8.6}{10.9} \approx 0.789
\]

Next, we find the probability that \(Z > 0.789\). From standard normal distribution tables or a calculator, the area to the left of \(Z = 0.789\) is approximately 0.7852. So, the probability of receiving more than 75 applications is:

\[
P(X > 75) = 1 - 0.7852 = 0.2148
\]

**b. At least 75 applications?**

"At least 75" means \(X \geq 75\), which is the same as "more than or equal to 75." Since the normal distribution is continuous, the probability for exactly 75 is zero. So, this is also:

\[
P(X \geq 75) = P(X > 75) = 0.2148
\]

**c. Between 65 and 75 applications inclusive?**

For this, we calculate the Z-scores for both 65 and 75 applications.

- For \(X = 65\):
\[
Z = \frac{65 - 66.4}{10.9} = \frac{-1.4}{10.9} \approx -0.128
\]

- For \(X = 75\), we already found the Z-score: \(Z = 0.789\).

Now, using the standard normal table:

- \(P(Z = -0.128)\) is approximately 0.4495.
- \(P(Z = 0.789)\) is approximately 0.7852.

The probability of receiving between 65 and 75 applications is:

\[
P(65 \leq X \leq 75) = P(-0.128 \leq Z \leq 0.789) = 0.7852 - 0.4495 = 0.3357
\]

### Summary of Probabilities:
- **a.** \(P(X > 75) = 0.2148\)
- **b.** \(P(X \geq 75) = 0.2148\)
- **c.** \(P(65 \leq X \leq 75) = 0.3357\)

Would you like further details or have any questions on these calculations?

### Related questions:
1. How do we calculate probabilities for other intervals, such as \(X > 80\)?
2. What does the Z-score represent in a normal distribution?
3. How can we compute probabilities for discrete variables in a normal approximation?
4. How does the normal distribution change when the mean or standard deviation changes?
5. What would happen to the probability if the standard deviation was reduced to 5?

### Tip:
When using Z-scores to find probabilities, remember that the standard normal table gives the probability to the left of the Z-score, so for values greater than \(Z\), subtract from 1.

The manager of a finance company knows from his experience that the number of loan applications received in his office during a week is a random variable with mean μ = 66.4 and standard deviation σ = 10.9. Assuming that the distribution of the number of loan applications is approximately normal, what is the probability that the office will receive in one week: a. More than 75 applications? b. At least 75 applications? c. Between 65 and 75 applications inclusive?

Calculate the probabilities for loan applications based on a normal distribution with a mean of 66.4 and standard deviation of 10.9. Learn how to find probabilities for more than 75 applications, at least 75 applications, and between 65 and 75 applications inclusive.

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