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.

To solve the given problem, we will go step by step to compute the requested statistics for the delivery time.

### Given Data:
- Delivery Times \( X \): 3, 4, 5, 6, 7 (hours)
- Corresponding Probabilities: 0.20, 0.35, 0.25, 0.15, 0.05

### a. Mean (Expected Value) of Delivery Time:
The mean \( E(X) \) or expected value is calculated as:
\[
E(X) = \sum [x_i \cdot P(x_i)]
\]
where \( x_i \) is each delivery time, and \( P(x_i) \) is the probability of each delivery time.

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

### b. Variance of Delivery Time:
The variance \( \text{Var}(X) \) is calculated using the formula:
\[
\text{Var}(X) = E(X^2) - [E(X)]^2
\]
First, we compute \( E(X^2) \):
\[
E(X^2) = \sum [x_i^2 \cdot P(x_i)]
\]
\[
E(X^2) = (3^2 \cdot 0.20) + (4^2 \cdot 0.35) + (5^2 \cdot 0.25) + (6^2 \cdot 0.15) + (7^2 \cdot 0.05)
\]
Then use the formula for variance.

### c. Standard Deviation of Delivery Time:
The standard deviation is simply the square root of the variance:
\[
\text{SD}(X) = \sqrt{\text{Var}(X)}
\]

### d. Probability that Delivery Time is Less than or Equal to 5 Hours:
\[
P(X \leq 5) = P(X = 3) + P(X = 4) + P(X = 5)
\]
\[
P(X \leq 5) = 0.20 + 0.35 + 0.25
\]

### e. Probability that Delivery Time is Greater than 6 Hours:
\[
P(X > 6) = P(X = 7)
\]
\[
P(X > 6) = 0.05
\]

Now, let's compute these values.### Results:

a. The **Mean (Expected Value)** of the delivery time is:
\[
E(X) = 4.5 \text{ hours}
\]

b. The **Variance** of the delivery time is:
\[
\text{Var}(X) = 1.25 \text{ hours}^2
\]

c. The **Standard Deviation** of the delivery time is:
\[
\text{SD}(X) = 1.12 \text{ hours}
\]

d. The **Probability that the delivery time is less than or equal to 5 hours** is:
\[
P(X \leq 5) = 0.80
\]

e. The **Probability that the delivery time is greater than 6 hours** is:
\[
P(X > 6) = 0.05
\]

Would you like further details on these computations or explanations?

### 5 Related Questions:
1. How would these results change if the probability distribution shifted?
2. What does the variance tell us about delivery times in this context?
3. How can we apply this distribution to forecast delivery time over multiple routes?
4. If the company wants to minimize variance, what kind of changes should be made?
5. How do these probabilities help in decision-making for optimal route planning?

### Tip:
For logistics optimization, reducing variance in delivery times can improve reliability and customer satisfaction.

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.

Learn how to calculate the mean (expected value), variance, and standard deviation for a logistics delivery time distribution. This problem involves applying key probability and statistical concepts and is suitable for students in Grades 11-12.

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