Performance on the test was generally strong. The mean was 93.7 and the standard deviation 8.2.

Here is the distribution of results.

Here is a display of the number of wrong responses, by question number. As you can see, items 14,16 were the most difficult.

Suppose a baseball player is really "a .300 hitter," and you model a season for this player with the binomial distribution based on 500 binomial trials (``at-bats'') with \(p = .300\) the probability of a hit in any at-bat. If the player remained absolutely consistent (i.e., \(p\) did not change), played 14 seasons, with exactly 500 at-bats per season, and his batting average (proportion of hits relative to at-bats) varied from season to season only through random variation, what would of the following would most closely match the expected value of his best single season batting average?

Each season is like 500 binomial trials. The proportion of hits is simply the number of hits divided by 500. We use the rbinom function to create 14 seasons, and then we create a function best.year to simply select the best season out of 14. We replicate this 100000 times to get an estimated expected value.

n <- 500
seasons <- 14
ba <- .300
best.year <- function()max(rbinom(seasons,n,ba))/n
mean(replicate(100000,best.year()))

## [1] 0.3352

#answer is about .3352

You have an IQ of 149. You walk into a room occupied by 5 other graduate students. Suppose that graduate student IQs are normally distributed with a mean of 130 and a standard deviation of 15, and the group of graduate students represents an independent, random sample. What is the probability that you have the highest IQ in the room?

For you to be the highest, all 5 of the people must be less than or equal to you. Call the probability that a single person will be less than or equal to 149 \(p\). Then the probability that all 5 will fail to reach 149 is \(p^5\).

Your.IQ <- 149
Mean <- 130
SD <- 15
p <- pnorm(Your.IQ,130,15) 
n <- 5
answer <- p^n
answer

## [1] 0.5819