Define the standardized (Z) score of a data point as the number of standard deviations it is away from the mean: \(Z = \frac{x - \mu}{\sigma}\).

Use the Z score

if the distribution is normal: to determine the percentile score of a data point (using technology or normal probability tables)

regardless of the shape of the distribution: to assess whether or not the particular observation is considered to be unusual (more than 2 standard deviations away from the mean)

Depending on the shape of the distribution determine whether the median would have a negative, positive, or 0 Z score.

Assess whether or not a distribution is nearly normal using the 68-95-99.7% rule or graphical methods such as a normal probability plot.

Reading: Section 4.1 of OpenIntro Statistics

Test yourself: True/False: In a right skewed distribution the Z score of the median is positive.

If X is a random variable that takes the value 1 with probability of success \(p\) and 0 with probability of success \(1-p\), then \(X\) is a Bernoulli random variable.

The geometric distribution is used to describe how many trials it takes to observe a success.

Define the probability of finding the first success in the \(n^{th}\) trial as \((1-p)^{n-1}p\).

\(\mu = \frac{1}{p}\)

\(\sigma^2 = \frac{1-p}{p^2}\)

\(\sigma = \sqrt{\frac{1-p}{p^2}}\)

Determine if a random variable is binomial using the four conditions:

The trials are independent.

The number of trials, n, is fixed.

Each trial outcome can be classified as a success or failure.

The probability of a success, p, is the same for each trial.

Calculate the number of possible scenarios for obtaining \(k\) successes in \(n\) trials using the choose function: \({n \choose k} = \frac{n!}{k!~(n - k)!}\).

Calculate probability of a given number of successes in a given number of trials using the binomial distribution: \(P(k = K) = \frac{n!}{k!~(n - k)!}~p^k~(1-p)^{(n - k)}\).

Calculate the expected number of successes in a given number of binomial trials \((\mu = np)\) and its standard deviation \((\sigma = \sqrt{np(1-p)})\).

When number of trials is sufficiently large (\(np \ge 10\) and \(n(1-p) \ge 10\)), use normal approximation to calculate binomial probabilities, and explain why this approach works.