September 18, 2019

Announcements

Presentations

Coin Tosses Revisited

coins <- sample(c(-1,1), 100, replace=TRUE)
plot(1:length(coins), cumsum(coins), type='l')
abline(h=0)

cumsum(coins)[length(coins)]
## [1] 14

Many Random Samples

samples <- rep(NA, 1000)
for(i in seq_along(samples)) {
    coins <- sample(c(-1,1), 100, replace=TRUE)
    samples[i] <- cumsum(coins)[length(coins)]
}
head(samples)
## [1]  0  8 -4 -6  2 -2

Histogram of Many Random Samples

hist(samples)

Properties of Distribution

(m.sam <- mean(samples))
## [1] 0.244
(s.sam <- sd(samples))
## [1] 10.05552

Properties of Distribution (cont.)

within1sd <- samples[samples >= m.sam - s.sam & samples <= m.sam + s.sam]
length(within1sd) / length(samples)
## [1] 0.679
within2sd <- samples[samples >= m.sam - 2 * s.sam & samples <= m.sam + 2* s.sam]
length(within2sd) / length(samples)
## [1] 0.959
within3sd <- samples[samples >= m.sam - 3 * s.sam & samples <= m.sam + 3 * s.sam]
length(within3sd) / length(samples)
## [1] 0.997

Standard Normal Distribution

\[ f\left( x|\mu ,\sigma \right) =\frac { 1 }{ \sigma \sqrt { 2\pi } } { e }^{ -\frac { { \left( x-\mu \right) }^{ 2 } }{ { 2\sigma }^{ 2 } } } \]

x <- seq(-4,4,length=200); y <- dnorm(x,mean=0, sd=1)
plot(x, y, type = "l", lwd = 2, xlim = c(-3.5,3.5), ylab='', xlab='z-score', yaxt='n')

Standard Normal Distribution

Standard Normal Distribution

Standard Normal Distribution

What’s the likelihood of ending with less than 15?

pnorm(15, mean=mean(samples), sd=sd(samples))
## [1] 0.9288734

What’s the likelihood of ending with more than 15?

1 - pnorm(15, mean=mean(samples), sd=sd(samples))
## [1] 0.07112657

Comparing Scores on Different Scales

SAT scores are distributed nearly normally with mean 1500 and standard deviation 300. ACT scores are distributed nearly normally with mean 21 and standard deviation 5. A college admissions officer wants to determine which of the two applicants scored better on their standardized test with respect to the other test takers: Pam, who earned an 1800 on her SAT, or Jim, who scored a 24 on his ACT?

Z-Scores

  • Z-scores are often called standard scores:

\[ Z = \frac{observation - mean}{SD} \]

  • Z-Scores have a mean = 0 and standard deviation = 1.

Converting Pam and Jim’s scores to z-scores:

\[ Z_{Pam} = \frac{1800 - 1500}{300} = 1 \]

\[ Z_{Jim} = \frac{24-21}{5} = 0.6 \]

Standard Normal Parameters