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.

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter1: Functions
Section1.2: The Least Square Line
Problem 1E
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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
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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