Question 1

Answer

What is the sample space?
The sample space is: [1,2,3,4,5,6]

What is P(2)
P(2) = \(\frac{1}{6}\)

What is P(8)?
P(8) = 0

What is the P({2, 4, 6})?
P({2,4,6}) = \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{1}{2}\)

* Let A = {1, 5, 6} and B = {1, 4, 6}
* What is P(A and B)?
P({1,6}) = \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{1}{3}\)

* What is P(A or B)?
You can do this by selecting the likelihood of choosing each 1, 4, 5, 6: = \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{2}{3}\)
or the formula of P(A or B) = P(A) + P(B) - P(A and B)
= P({1,5,6}) = \(\frac{3}{6}\) + P({1,4,6}) = \(\frac{3}{6}\) - P({1,4,5,6}) = \(\frac{2}{6}\) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Question 2

Answer

P(sums = 11)
Total number of outcomes = \(6^2\) There are only two ways to get 11, 5 + 6 or 6 + 5, therefore, P(sums = 11) = \(\frac{2}{36}\) = \(\frac{1}{18}\)

P(sums = 12)
Total number of outcomes = \(6^2\)
You can get a sum of 12 by 6+6 only, therefore
P(sums = 12) = \(\frac{1}{36}\)

Question 3

Answer

Good = \(\frac{95}{100}\)
Passes = (1 - unusable)*100 = \(\frac{98}{100}\)
P(Good) = \(\frac{\frac{95}{100}}{\frac{98}{100}}\) = \(\frac{95}{98}\)
96.93% likelihood of being good if passed

Question 4

Answer

Probablity of guessing the 2 defective parts in the first 2 picks
P(of getting a defective part in the first pick) = \(\frac{2}{10}\)
Note that there is no replacement here
P(of getting a defective part in the second pick) = \(\frac{1}{9}\)
P(getting both defective parts in the first two picks) = \(\frac{2}{10}\) * \(\frac{1}{9}\) = \(\frac{1}{45}\)
The probability is 2.22% of getting both defective parts in the first two picks.

Question 5

Answer

## The mean of this random sample is, 47.45255
## The standard deviation of this random sample is, 5.392391

## The mean of this random sample is, 45.03581
## The standard deviation of this random sample is, 6.797605


Based on the law of large numbers, the average number of results from a large series of trials should be close to the expected value and becomes incrementally closer to the expected value as it becomes bigger; also better representing a normal distribution.

Code

set.seed(365)
random = rnorm(n = 10, mean = 45, sd = 7)
rand.mean = mean(random)
cat("The mean of this random sample is,", rand.mean)

rand.sd = sd(random)
cat("The standard deviation of this random sample is,", rand.sd)

hist(random, freq = FALSE, main = "Histogram of Random Sample Size of 10", xlab = "Random Numbers", col = "#3EC0E6", xlim = c(20, 70) ) 
x =  seq(from = 20, to = 70, by = 0.01)  
lines(x, dnorm(x, 45, 7))

set.seed(365)
random.1000 = rnorm(n = 1000, mean = 45, sd = 7)
rand.mean.1000 = mean(random.1000)
cat("The mean of this random sample is,", rand.mean.1000)

rand.sd.1000 = sd(random.1000)
cat("The standard deviation of this random sample is,", rand.sd.1000)

hist(random.1000, freq = FALSE, main = "Histogram of Random Sample Size of 1000", xlab = "Random Numbers", col = "#1770E1", xlim = c(20, 70) ) 
x =  seq(from = 20, to = 70, by = 0.01)  
lines(x, dnorm(x, 45, 7))

Question 6

Answer

What is the probability that a randomly selected individual has 11 years of education or more? 

## [1] 69.15

What is the probability that a randomly selected individual has between 16 and 20 years of education (an advanced degree)? 

## [1] 18.66

Code

round(pnorm(11, mean = 13, sd = 4, lower.tail = FALSE),digits = 4)*100


round(pnorm(20, mean = 13, sd = 4, lower.tail = TRUE) - 
pnorm(16, mean = 13, sd = 4, lower.tail = TRUE), digits = 4)*100