Let X₁, X2, X196 be a random sample of 196 observations from the distribution of X. Let X be the sample mean. Use R to determine the following: a) Find the approximate probability P(X> 1.1)| b) What is the approximate probability that X₁ + X₂ + ... +X196 >5.3 c) Copy your R script for the above into the text box here.
Let X₁, X2, X196 be a random sample of 196 observations from the distribution of X. Let X be the sample mean. Use R to determine the following: a) Find the approximate probability P(X> 1.1)| b) What is the approximate probability that X₁ + X₂ + ... +X196 >5.3 c) Copy your R script for the above into the text box here.
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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![### Problem Statement:
Suppose \( X \) is a random variable with an expected value \( \mu = 0 \) and standard deviation \( \sigma = 4\sqrt{5} \).
Let \( X_1, X_2, \ldots, X_{196} \) be a random sample of 196 observations from the distribution of \( X \).
Let \( \bar{X} \) be the sample mean. Use R to determine the following:
a) Find the approximate probability \( P(\bar{X} > 1.1) \).
[Text box for answer]
b) What is the approximate probability that \( X_1 + X_2 + \ldots + X_{196} > 5.3 \)?
[Text box for answer]
c) Copy your R script for the above into the text box here.
[Text box for R script]
### Explanation:
The problem involves computing probabilities for sample mean and sum of a normally distributed random variable using R programming. The central limit theorem can be useful here, given the large sample size.
To solve part (a), one may use the fact that for a large sample size, the sampling distribution of the sample mean \( \bar{X} \) is approximately normal. Similarly, for part (b), the sum of the normally distributed random variables can be considered.
Please note: To solve these problems, you'd likely simulate this in R or derive the results through normal distribution properties, making use of mean and standard deviation.
### R Code for Part (c):
```r
# Part (a)
mu <- 0
sigma <- 4 * sqrt(5)
n <- 196
x_bar <- 1.1
# Standard deviation of sample mean (Standard Error)
sigma_x_bar <- sigma / sqrt(n)
# Approximate probability P(X_bar > 1.1)
prob_a <- 1 - pnorm(x_bar, mean = mu, sd = sigma_x_bar)
print(prob_a)
# Part (b)
sum_value <- 5.3
mu_sum <- mu * n
sigma_sum <- sigma * sqrt(n)
# Approximate probability P(X1 + X2 + ... + X196 > 5.3)
prob_b <- 1 - pnorm(sum_value, mean = mu_sum, sd = sigma_sum)
print(prob_b)
```
This script calculates the probabilities using the properties of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa61d77b8-3322-4869-a508-5eb685631aec%2F26255c5f-145c-4c3a-9303-95b00aefbe0c%2Ffhg91l_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Suppose \( X \) is a random variable with an expected value \( \mu = 0 \) and standard deviation \( \sigma = 4\sqrt{5} \).
Let \( X_1, X_2, \ldots, X_{196} \) be a random sample of 196 observations from the distribution of \( X \).
Let \( \bar{X} \) be the sample mean. Use R to determine the following:
a) Find the approximate probability \( P(\bar{X} > 1.1) \).
[Text box for answer]
b) What is the approximate probability that \( X_1 + X_2 + \ldots + X_{196} > 5.3 \)?
[Text box for answer]
c) Copy your R script for the above into the text box here.
[Text box for R script]
### Explanation:
The problem involves computing probabilities for sample mean and sum of a normally distributed random variable using R programming. The central limit theorem can be useful here, given the large sample size.
To solve part (a), one may use the fact that for a large sample size, the sampling distribution of the sample mean \( \bar{X} \) is approximately normal. Similarly, for part (b), the sum of the normally distributed random variables can be considered.
Please note: To solve these problems, you'd likely simulate this in R or derive the results through normal distribution properties, making use of mean and standard deviation.
### R Code for Part (c):
```r
# Part (a)
mu <- 0
sigma <- 4 * sqrt(5)
n <- 196
x_bar <- 1.1
# Standard deviation of sample mean (Standard Error)
sigma_x_bar <- sigma / sqrt(n)
# Approximate probability P(X_bar > 1.1)
prob_a <- 1 - pnorm(x_bar, mean = mu, sd = sigma_x_bar)
print(prob_a)
# Part (b)
sum_value <- 5.3
mu_sum <- mu * n
sigma_sum <- sigma * sqrt(n)
# Approximate probability P(X1 + X2 + ... + X196 > 5.3)
prob_b <- 1 - pnorm(sum_value, mean = mu_sum, sd = sigma_sum)
print(prob_b)
```
This script calculates the probabilities using the properties of
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