Alec Gremer's Profile

  Module # 6 Assignment Part A. a) To compute the mean of the population, you sum up all the values and divide by the total number of values: Population mean = (8 + 14 + 16 + 10 + 11) / 5 = 59 / 5 = 11.8 b) You can use the sample() function in R to randomly select a sample of size 2 from the population, such as this one: population <- c(8, 14, 16, 10, 11) sample_size <- 2 sample_data <- sample(population, size = sample_size) c) To compute the mean and standard deviation of the sample: members <- c(3, 5, 2, 1) sample_size <- 2 sample_data <- sample(members, size = sample_size) sample_data Sample Mean = 9.5 Sample Standard Deviation = 2.12 d) The sample mean (9.5) is lower than the population mean (11.8). This suggests that, on average, the sample is purchasing fewer ice creams during the academic year compared to the entire population. Part B. Does the sample proportion p have approximately a normal distribution? The sample proportion, denoted as p̂ (p-hat), has an...

 Module #4 Probability Theory A1. Event A: P(A) = 10  A2. Event B: P(B) = 20  A3. Event A1: P(A1) = 20 A4. Event B1: P(B1) = 40    B Event B1: True. The probability that it will rain on the day of Jane's wedding, given the weatherman's forecast for rain, is approximately 11.1%. B Event B2: The result of successfully applying Bayes' theorem to the given data makes the answer accurate because it is the outcome that was attained. The Bayes theorem is a formula for revising probability in light of fresh information or evidence, in this case, a weather forecast. The revised likelihood that it will rain on the wedding day is determined using the prior likelihood of rain (P(A1)) as well as the conditional probabilities of the weatherman's forecast (P(B | A1) and P(B | A2).   C: dbinom(X, size=N, prob=P) dbinom(10, size=10, prob=0.2)   0.1073742   With this approach, the likelihood of successfully operating on 10 patients is roughly 0.1074,...

Data Set Analysis Set #1: 10, 2, 3, 2, 4, 2, 5 Set #2: 20, 12, 13, 12, 14, 12, 15 1) Central Tendency: Mean : Set #1: (10 + 2 + 3 + 2 + 4 + 2 + 5) / 7 = 28 / 7 ≈ 4 Set #2: (20 + 12 + 13 + 12 + 14 + 12 + 15) / 7 = 98 / 7 ≈ 14 Median: Set #1: Median = 3  Set #2: Median = 14  Mode: Set #1: Mode = 2 Set #2: Mode = 12 2) Variation: Range: Set #1: Range = 8 Set #2: Range = 8 Interquartile Range (IQR): Set #1: IQR = Q3 - Q1 Set #2: IQR = Q3 - Q1 Variance: Set #1: Variance = ~6.33  Set #2: Variance = ~9.33 Standard Deviation: Set #1: Standard Deviation = ~2.52 Set #2: Standard Deviation = ~3.06 Comparison of Central Tendency: Comparing Set #1 and Set #2, the mean, median, and mode values in Set #2 are typically higher. This shows that the average value in Set #2 is greater. Variation: Given that the range is the same for both sets, the spread of data from minimum to maximum is also identical. The interquartile range (IQR) of Set #2 is bigger than that of Set #1, indicating that S...

Calculating Vector Mean The R script creates the myMean function, which computes the average of the assignment2 vector. The sum(assignment2) code calculates the total of each component in the vector assignment2. Using the formula, length(assignment2), the vector's length, or the number of assignment2's elements, is determined. The mean is then calculated by dividing the sum of the elements by the length of the vector. The assignment2 vector's mean value is around 18.66667.