Question 2 Let X be a discrete random variable with probability mass function given by 1 2 3 p(x) 1/4 1/2 1/8 1/8 a) Estimate P(X < 1). Answer: в «- 1000 x <- replicate(B, sample( c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8) ) mean (x <= 1) ## [1] 0.76 # answer (1/4) + (1/2) ## [1] 0.75 b) Find P(X <1|X < 2). Answer: B <- 1000 x <- replicate (B, sample( c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8) ), y <- replicate(B, sample( c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8) ) mean (x <= 1)/mean (y <= 2)
Question 2 Let X be a discrete random variable with probability mass function given by 1 2 3 p(x) 1/4 1/2 1/8 1/8 a) Estimate P(X < 1). Answer: в «- 1000 x <- replicate(B, sample( c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8) ) mean (x <= 1) ## [1] 0.76 # answer (1/4) + (1/2) ## [1] 0.75 b) Find P(X <1|X < 2). Answer: B <- 1000 x <- replicate (B, sample( c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8) ), y <- replicate(B, sample( c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8) ) mean (x <= 1)/mean (y <= 2)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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This is a Statistics problem. Please code in R if needed.
The problem I'm having difficulty understanding is shown below:
![Question 2
Let \( X \) be a discrete random variable with probability mass function given by
\[
\begin{array}{c|cccc}
x & 0 & 1 & 2 & 3 \\
\hline
p(x) & 1/4 & 1/2 & 1/8 & 1/8 \\
\end{array}
\]
a) Estimate \( P(X \leq 1) \).
**Answer:**
```R
B <- 1000
x <- replicate(B,
sample(c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8))
)
mean(x <= 1)
```
Output:
```
## [1] 0.76
```
**Theoretical Calculation:**
\[
P(X \leq 1) = (1/4) + (1/2)
\]
Output:
```
## [1] 0.75
```
b) Find \( P(X \leq 1 \mid X \leq 2) \).
**Answer:**
```R
B <- 1000
x <- replicate(B,
sample(c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8))
)
y <- replicate(B,
sample(c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8))
)
mean(x <= 1)/mean(y <= 2)
```
Output:
```
## [1] 0.8963134
```
**Theoretical Calculation:**
\[
P(X \leq 1 \mid X \leq 2) = \frac{(1/4 + 1/2)}{(1/4 + 1/2 + 1/8)}
\]
Output:
```
## [1] 0.8571429
```
**Explanation:**
This section demonstrates estimating probabilities for the discrete random variable \( X \) using simulation and theoretical calculations. Part (a) estimates the probability of \( X \) being less than or equal to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb9da028-c1e8-4fb4-9df2-7f4287e8030e%2F6aa38585-6abe-4e5a-9f58-99702569fb0e%2Fsw45pp8_processed.png&w=3840&q=75)
Transcribed Image Text:Question 2
Let \( X \) be a discrete random variable with probability mass function given by
\[
\begin{array}{c|cccc}
x & 0 & 1 & 2 & 3 \\
\hline
p(x) & 1/4 & 1/2 & 1/8 & 1/8 \\
\end{array}
\]
a) Estimate \( P(X \leq 1) \).
**Answer:**
```R
B <- 1000
x <- replicate(B,
sample(c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8))
)
mean(x <= 1)
```
Output:
```
## [1] 0.76
```
**Theoretical Calculation:**
\[
P(X \leq 1) = (1/4) + (1/2)
\]
Output:
```
## [1] 0.75
```
b) Find \( P(X \leq 1 \mid X \leq 2) \).
**Answer:**
```R
B <- 1000
x <- replicate(B,
sample(c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8))
)
y <- replicate(B,
sample(c(0, 1, 2, 3), 1, prob = c(1/4, 1/2, 1/8, 1/8))
)
mean(x <= 1)/mean(y <= 2)
```
Output:
```
## [1] 0.8963134
```
**Theoretical Calculation:**
\[
P(X \leq 1 \mid X \leq 2) = \frac{(1/4 + 1/2)}{(1/4 + 1/2 + 1/8)}
\]
Output:
```
## [1] 0.8571429
```
**Explanation:**
This section demonstrates estimating probabilities for the discrete random variable \( X \) using simulation and theoretical calculations. Part (a) estimates the probability of \( X \) being less than or equal to
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