Suppose we choose independently 25 numbers at random (uniform density) from the interval [0, 20]. We are interested in finding the normal densities that approximate the densities of their sum \(S_{25}\), their standardized sum \(S^*_{25}\), and their average \(A_{25}\).
We are interested in approximating the distributions of the sum \(S_{25}\), the standardized sum \(S^*_{25}\), and the average \(A_{25}\) of 25 independently chosen random numbers from the interval [0, 20] using normal densities.
The sum \(S_{25}\) is the total of 25 randomly chosen numbers from a uniform distribution between 0 and 20.
Therefore, the normal density approximating \(S_{25}\) is \(N(250, 28.87)\).
The standardized sum \(S^*_{25}\) is the sum \(S_{25}\) minus its mean, divided by its standard deviation.
Hence, the normal density that approximates \(S^*_{25}\) is \(N(0, 1)\).
The average \(A_{25}\) is the mean of 25 uniformly distributed random numbers between 0 and 20.
The normal density that approximates \(A_{25}\) is \(N(10, 1.15)\).
By using the Central Limit Theorem, these normal densities provide a good approximation for the distributions of the sum, standardized sum, and average of the 25 randomly chosen numbers.
The sum \(S_{25}\) is the sum of 25 independent and identically distributed random variables from a uniform distribution over [0, 20].
\[ E[U(0, 20)] = \frac{0 + 20}{2} = 10 \]
\[ E[S_{25}] = 25 \times E[U(0, 20)] = 25 \times 10 = 250 \]
\[ \text{Var}(U(0, 20)) = \frac{(20 - 0)^2}{12} = \frac{400}{12} \approx 33.33 \]
\[ \text{Var}(S_{25}) = 25 \times \text{Var}(U(0, 20)) = 25 \times 33.33 \approx 833.33 \]
\[ S_{25} \sim N(250, \sqrt{833.33}) \]
The standardized sum, \(S^*_{25}\), is obtained by subtracting the expected value (mean) and dividing by the standard deviation. The formula is given by:
\[ S^*_{25} = \frac{S_{25} - E[S_{25}]}{\sqrt{\text{Var}(S_{25})}} \]
For \(S_{25}\), which is the sum of 25 independent random variables uniformly distributed from 0 to 20, we have:
Thus, when we standardize \(S_{25}\):
This is because the standardization process shifts the mean to 0 and scales the distribution to have a unit variance.
The average \(A_{25}\) is the sum \(S_{25}\) divided by 25.
\[ E[A_{25}] = \frac{E[S_{25}]}{25} = \frac{250}{25} = 10 \]
\[ \text{Var}(A_{25}) = \frac{\text{Var}(S_{25})}{25^2} = \frac{833.33}{625} \approx 1.33 \]
\[ A_{25} \sim N(10, \sqrt{1.33}) \]
Below is the r code to deduce these normal approximations for \(S_{25}\), \(S^*_{25}\), and \(A_{25}\).
# Set seed for reproducibility
set.seed(123)
# Generate 25 random numbers from U(0, 20)
random_numbers <- runif(25, min = 0, max = 20)
# Sum S_25
S_25 <- sum(random_numbers)
# Standardized Sum S*_25
mean_S_25 <- mean(random_numbers) * 25
std_dev_S_25 <- sqrt(25 * (20^2 / 12))
S_star_25 <- (S_25 - mean_S_25) / std_dev_S_25
# Average A_25
A_25 <- mean(random_numbers)
# Display results
cat("Sum S_25:", S_25, "\n")
## Sum S_25: 297.7799
cat("Standardized Sum S*_25:", S_star_25, "\n")
## Standardized Sum S*_25: 0
cat("Average A_25:", A_25, "\n")
## Average A_25: 11.91119