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.