Exponential Distribution Mean, Coin Flip Standard Deviation and Expected Value

Math Problem Statement

A call center technical specialist spends varying amount of time in each call to resolve the concern. The time spent in each call is modeled using the following distribution: X~ Exp (0.2).

What is the mean?

Question 1 Answer a.

1 b.

5 c.

2 Question 2 Not yet answered Marked out of 1.00 Flag question Question text

A coin is flipped once. Let X the random variable representing the two possible outcomes with the following probability distribution:

P(x)

P(x=head) = 0.5

P(x=tail) = 0.5

What is the standard deviation of X?

Question 2 Answer a.

0.50

0.00 c.

0.707

Question 3 Not yet answered Marked out of 1.00 Flag question Question text

A coin is flipped once. Let X the random variable representing the two possible outcomes with the following probability distribution:

P(x)

P(x=head) = 0.5

P(x=tail) = 0.5

What is the expected value of X?

Question 3 Answer a.

0.5

0 c.

0.25

Solution

Question 1:

Given that the time spent on each call follows an exponential distribution, $X \sim \text{Exp}(\lambda = 0.2)$ , the mean $\mu$ for an exponential distribution is given by:

$\mu = \frac{1}{\lambda}$

Substituting the given value of $\lambda = 0.2$ :

$\mu = \frac{1}{0.2} = 5$

Answer: The mean is $5$ . Therefore, the correct option is b. 5.

Question 2:

We are given a discrete random variable $X$ , representing the outcome of a fair coin flip with the following probability distribution:

0 & \text{with probability } P(x = 0) = 0.5 \\ 1 & \text{with probability } P(x = 1) = 0.5 \end{cases}$$ The **standard deviation** is the square root of the variance. First, we need to calculate the **variance** $$\sigma^2$$. 1. **Expected Value (Mean) $$\mu$$**: $$\mu = E(X) = (0 \times 0.5) + (1 \times 0.5) = 0 + 0.5 = 0.5$$ 2. **Variance** $$\sigma^2$$: $$\sigma^2 = E(X^2) - (E(X))^2$$ We already know $$E(X) = 0.5$$. Now, calculate $$E(X^2)$$: $$E(X^2) = (0^2 \times 0.5) + (1^2 \times 0.5) = 0 + 0.5 = 0.5$$ Now, compute the variance: $$\sigma^2 = 0.5 - (0.5)^2 = 0.5 - 0.25 = 0.25$$ 3. **Standard Deviation** $$\sigma$$: $$\sigma = \sqrt{0.25} = 0.5$$ **Answer:** The standard deviation is $$0.5$$. Therefore, the correct option is **a. 0.50**. --- ### Question 3: We are again given the same random variable $$X$$ with the distribution: $$X = \begin{cases} 0 & \text{with probability } P(x = 0) = 0.5 \\ 1 & \text{with probability } P(x = 1) = 0.5 \end{cases}$$ The **expected value** $$E(X)$$ (mean) is calculated as: $$E(X) = (0 \times 0.5) + (1 \times 0.5) = 0 + 0.5 = 0.5$$ **Answer:** The expected value is $$0.5$$. Therefore, the correct option is **a. 0.5**. --- ### Summary of Answers: 1. **Mean**: b. 5 2. **Standard Deviation**: a. 0.50 3. **Expected Value**: a. 0.5 --- Would you like further clarification on any of the solutions, or do you have other questions? Here are 5 related questions you might find helpful: 1. How would the mean change if the rate parameter $$\lambda$$ in an exponential distribution is increased? 2. What happens to the variance of a coin flip when the probabilities of the outcomes are not equal (e.g., 0.7 for heads and 0.3 for tails)? 3. How can you compute the mean and standard deviation for a binomial distribution? 4. What is the relationship between the mean, variance, and standard deviation of a normal distribution? 5. Can the expected value ever be negative for a discrete random variable? Why or why not? **Tip**: For discrete distributions, you can use the formula $$\mu = E(X) = \sum x_i \cdot P(x_i)$$ to calculate the expected value, where $$x_i$$ are the possible outcomes and $$P(x_i)$$ are their associated probabilities.

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Math Problem Analysis

Mathematical Concepts

Exponential Distribution
Probability
Expected Value
Standard Deviation

Formulas

Mean of Exponential Distribution: μ = 1/λ
Expected Value: E(X) = Σ x_i * P(x_i)
Variance: σ^2 = E(X^2) - (E(X))^2
Standard Deviation: σ = √Variance

Theorems

Properties of Exponential Distribution
Discrete Probability Distributions
Variance and Standard Deviation Formulas

Suitable Grade Level

Grades 9-12

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