Math Problem Statement

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.

Solution

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

Given Data:

  • Delivery Times XX: 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)E(X) or expected value is calculated as: E(X)=[xiP(xi)]E(X) = \sum [x_i \cdot P(x_i)] where xix_i is each delivery time, and P(xi)P(x_i) is the probability of each delivery time.

E(X)=(30.20)+(40.35)+(50.25)+(60.15)+(70.05)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 Var(X)\text{Var}(X) is calculated using the formula: Var(X)=E(X2)[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2 First, we compute E(X2)E(X^2): E(X2)=[xi2P(xi)]E(X^2) = \sum [x_i^2 \cdot P(x_i)] E(X2)=(320.20)+(420.35)+(520.25)+(620.15)+(720.05)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: SD(X)=Var(X)\text{SD}(X) = \sqrt{\text{Var}(X)}

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

P(X5)=P(X=3)+P(X=4)+P(X=5)P(X \leq 5) = P(X = 3) + P(X = 4) + P(X = 5) P(X5)=0.20+0.35+0.25P(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) = P(X = 7) P(X>6)=0.05P(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 hoursE(X) = 4.5 \text{ hours}

b. The Variance of the delivery time is: Var(X)=1.25 hours2\text{Var}(X) = 1.25 \text{ hours}^2

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

d. The Probability that the delivery time is less than or equal to 5 hours is: P(X5)=0.80P(X \leq 5) = 0.80

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

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Variance
Standard Deviation

Formulas

Mean (Expected Value) E(X) = ∑[xi * P(xi)]
Variance Var(X) = E(X^2) - [E(X)]^2
Standard Deviation SD(X) = √Var(X)
Probability P(X ≤ 5) = P(X = 3) + P(X = 4) + P(X = 5)
Probability P(X > 6) = P(X = 7)

Theorems

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Suitable Grade Level

Grades 10-12