A logistics company is interested in modeling the delivery time for a certain route. Let the random variable X represent the delivery time in hours. The probability distribution for X is given below: Delivery Time (hours) 3 4 5 6 7 Probability 0.20 0.35 0.25 0.15 0.05 a. Calculate the Mean (Expected Value) of the Delivery Time. b. Calculate the Variance of the Delivery Time. c. Calculate the Standard Deviation of the Delivery Time. d. Find the Probability that Delivery Time is Less than or Equal to 5 Hours. e. Find the Probability that Delivery Time is Greater than 6 Hours.

Let's address each part of the problem step by step.

### Given Data:
- Delivery Time (hours) \(X\): 3, 4, 5, 6, 7
- Probability of each delivery time:
  \[
  P(X = 3) = 0.20, \quad P(X = 4) = 0.35, \quad P(X = 5) = 0.25, \quad P(X = 6) = 0.15, \quad P(X = 7) = 0.05
  \]

### a. **Mean (Expected Value) of the Delivery Time**

The mean or expected value \( E(X) \) is calculated using the formula:
\[
E(X) = \sum [x_i \cdot P(x_i)]
\]
Where \( x_i \) are the values of the random variable (delivery times) and \( P(x_i) \) are the corresponding probabilities.

\[
E(X) = (3 \times 0.20) + (4 \times 0.35) + (5 \times 0.25) + (6 \times 0.15) + (7 \times 0.05)
\]

Let's calculate this value.

### b. **Variance of the Delivery Time**

The variance \( \text{Var}(X) \) is calculated using the formula:
\[
\text{Var}(X) = \sum [(x_i - E(X))^2 \cdot P(x_i)]
\]
We first need to find the mean \( E(X) \) (from part a) and then apply the formula.

### c. **Standard Deviation of the Delivery Time**

The standard deviation \( \sigma \) is simply the square root of the variance:
\[
\sigma = \sqrt{\text{Var}(X)}
\]

### d. **Probability that Delivery Time is Less than or Equal to 5 Hours**

We need to sum the probabilities where \( X \leq 5 \). This involves summing the probabilities for \( X = 3 \), \( X = 4 \), and \( X = 5 \).

\[
P(X \leq 5) = P(X = 3) + P(X = 4) + P(X = 5)
\]

### e. **Probability that Delivery Time is Greater than 6 Hours**

This involves summing the probabilities for \( X > 6 \), i.e., for \( X = 7 \).

\[
P(X > 6) = P(X = 7)
\]

I'll now compute each of these values.### Results:
a. **Mean (Expected Value) of the Delivery Time**: \( E(X) = 4.5 \) hours

b. **Variance of the Delivery Time**: \( \text{Var}(X) = 1.25 \)

c. **Standard Deviation of the Delivery Time**: \( \sigma = 1.12 \) hours (approximately)

d. **Probability that Delivery Time is Less than or Equal to 5 Hours**: \( P(X \leq 5) = 0.80 \)

e. **Probability that Delivery Time is Greater than 6 Hours**: \( P(X > 6) = 0.05 \)

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

Here are 5 related questions for practice:
1. What is the probability that the delivery time is exactly 6 hours?
2. What is the probability that the delivery time is greater than 4 hours?
3. How would the mean change if the delivery time of 3 hours had a higher probability?
4. What is the cumulative probability for delivery times less than 7 hours?
5. What is the z-score for a delivery time of 7 hours?

**Tip:** Variance measures the spread of the distribution, while the standard deviation provides this spread in the same units as the original data.

This problem focuses on calculating the mean, variance, and standard deviation for a given discrete probability distribution of delivery times. You will also explore how to find probabilities for specific conditions, such as delivery time being less than or equal to 5 hours and greater than 6 hours. Suitable for high school students in Grades 10-12.

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