4. A district court judge presides over many court cases each day. Let the random variable X= the number of cases on a randomly selected day in which she declares the defendant guilty. The distribution for the random variable X is given below: # of guilty 1 3 4 5 verdicts Probability 0.05 0.10 0.20 0.25 0.25 0.15 If we selected two days at random, what is the mean and standard deviation of the total number of defendants declared guilty by the judge over those two days? A. µ=3, o=1.90 B. µ=3, o=1.38 C. µ=6, o=1.95 D.) u=6, 0=3.80
4. A district court judge presides over many court cases each day. Let the random variable X= the number of cases on a randomly selected day in which she declares the defendant guilty. The distribution for the random variable X is given below: # of guilty 1 3 4 5 verdicts Probability 0.05 0.10 0.20 0.25 0.25 0.15 If we selected two days at random, what is the mean and standard deviation of the total number of defendants declared guilty by the judge over those two days? A. µ=3, o=1.90 B. µ=3, o=1.38 C. µ=6, o=1.95 D.) u=6, 0=3.80
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![**Problem Statement:**
A district court judge presides over many court cases each day. Let the random variable \(X\) represent the number of cases on a randomly selected day in which the judge declares the defendant guilty. The probability distribution for the random variable \(X\) is given below:
| # of guilty verdicts | 0 | 1 | 2 | 3 | 4 | 5 |
|----------------------|-----|-----|-----|-----|-----|-----|
| Probability | 0.05| 0.10| 0.20| 0.25| 0.25| 0.15|
**Question:**
If we selected two days at random, what is the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the total number of defendants declared guilty by the judge over those two days?
**Options:**
A. \(\mu=3, \sigma=1.90\)
B. \(\mu=3, \sigma=1.38\)
C. \(\mu=6, \sigma=1.95\)
D. \(\mu=6, \sigma=3.80\) (correct answer)
**Explanation:**
To determine the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the total number of defendants declared guilty over two days, you would add the means and variances of the single-day distributions:
1. **Calculate the mean for one day:**
\[
\mu_X = \sum (x_i \times P(x_i))
\]
\[
\mu_X = (0 \times 0.05) + (1 \times 0.10) + (2 \times 0.20) + (3 \times 0.25) + (4 \times 0.25) + (5 \times 0.15)
\]
\[
\mu_X = 3
\]
2. **Calculate the variance for one day:**
\[
\sigma_X^2 = \sum ((x_i - \mu_X)^2 \times P(x_i))
\]
\[
\sigma_X^2 = [(0-3)^2 \times 0.05] + [(1-3)^2 \times 0.10]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F734b2aa5-0d10-44d3-8b6f-53aad688fd5a%2Fcb62ae6c-41bf-4ab7-b84f-8727d6339217%2Fdis414c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A district court judge presides over many court cases each day. Let the random variable \(X\) represent the number of cases on a randomly selected day in which the judge declares the defendant guilty. The probability distribution for the random variable \(X\) is given below:
| # of guilty verdicts | 0 | 1 | 2 | 3 | 4 | 5 |
|----------------------|-----|-----|-----|-----|-----|-----|
| Probability | 0.05| 0.10| 0.20| 0.25| 0.25| 0.15|
**Question:**
If we selected two days at random, what is the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the total number of defendants declared guilty by the judge over those two days?
**Options:**
A. \(\mu=3, \sigma=1.90\)
B. \(\mu=3, \sigma=1.38\)
C. \(\mu=6, \sigma=1.95\)
D. \(\mu=6, \sigma=3.80\) (correct answer)
**Explanation:**
To determine the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the total number of defendants declared guilty over two days, you would add the means and variances of the single-day distributions:
1. **Calculate the mean for one day:**
\[
\mu_X = \sum (x_i \times P(x_i))
\]
\[
\mu_X = (0 \times 0.05) + (1 \times 0.10) + (2 \times 0.20) + (3 \times 0.25) + (4 \times 0.25) + (5 \times 0.15)
\]
\[
\mu_X = 3
\]
2. **Calculate the variance for one day:**
\[
\sigma_X^2 = \sum ((x_i - \mu_X)^2 \times P(x_i))
\]
\[
\sigma_X^2 = [(0-3)^2 \times 0.05] + [(1-3)^2 \times 0.10]
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