Please assist with all questions and add R script as well. ### 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 <- 0sigma <- 4 * sqrt(5)n <- 196x_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 <- 5.3mu_sum <- mu * nsigma_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

Suppose X is a random variable with expected value and standard deviation Let, X1, X2,....,X196 be…

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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Question

Please assist with all questions and add R script as well.

### 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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