Probability Distributions, Normal Approximation, and Sampling Distributions: Study Notes

Probability Distributions and the Normal Approximation

Binomial Distribution

The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success.

• Mean (Expected Value): $\mu_X = np$ Where n is the number of trials and p is the probability of success.

• Standard Deviation: $\sigma_X = \sqrt{np(1-p)}$

• Shape: The binomial distribution is approximately bell-shaped (normal) if $np \geq 10$ and $n(1-p) \geq 10$.

Example: For $n = 200$, $p = 0.30$:

• Mean: $\mu_X = 200 \times 0.30 = 60$

• Standard Deviation: $\sigma_X = \sqrt{200 \times 0.30 \times 0.70} \approx 6.48$

• Since $np = 60$ and $n(1-p) = 140$, the distribution is bell-shaped.

Normal Approximation to the Binomial

When the binomial distribution is approximately normal, probabilities can be estimated using the normal distribution.

• Draw a normal curve centered at $\mu_X$ with spread $\sigma_X$.

• Use continuity correction when approximating discrete binomial probabilities with the normal distribution.

Calculating Binomial Probabilities

• At least k successes: $P(X \geq k) = 1 - \text{binomcdf}(n, p, k-1)$

• At most k successes: $P(X \leq k) = \text{binomcdf}(n, p, k)$

Example: Probability that at least 75 papers are missing a name in $n=200$, $p=0.30$:

• $P(X \geq 75) = 1 - \text{binomcdf}(200, 0.30, 74) = 0.0138$

Normal Distribution and Z-Scores

Standard Normal Distribution

The standard normal distribution is a normal distribution with mean 0 and standard deviation 1. Any normal variable $X$ can be standardized:

• Z-score: $z = \frac{x - \mu}{\sigma}$

Z-scores measure how many standard deviations a value is from the mean.

Finding Probabilities and Percentiles

• Probability above a value: $P(X > x) = P(z > z_0)$

• Probability below a value: $P(X < x) = P(z < z_0)$

• Percentiles: To find the value corresponding to the $k$th percentile, use $x = \mu + z_k \sigma$ where $z_k$ is the z-score for the percentile.

Example: For $\mu = 72.6$, $\sigma = 10.6$, probability of scoring above 70:

• $z = \frac{70 - 72.6}{10.6} = -0.25$

• $P(X > 70) = P(z > -0.25) = 0.5987$

Unusual Values

• A value is considered unusual if its probability is less than 0.05.

• Example: If $P(X < 55) = 0.0485$, then 55 is an unusual score.

Middle 50% of a Normal Distribution

• Find z-scores for the 25th and 75th percentiles ($z_{0.25} = -0.674$, $z_{0.75} = 0.674$).

• Calculate $x_1 = \mu + z_{0.25}\sigma$, $x_2 = \mu + z_{0.75}\sigma$.

Applications: Waiting Times and Probability Plots

Normal Probability Plots

A normal probability plot is used to assess if a data set is approximately normally distributed. If the points fall roughly along a straight line, the data are considered normal.

Example: Waiting Times

• Given $\mu = 6.4$ min, $\sigma = 1.5$ min.

• Probability a person waits between 5 and 8 minutes:

  • $z_1 = \frac{5 - 6.4}{1.5} = -0.93$

  • $z_2 = \frac{8 - 6.4}{1.5} = 1.07$

  • $P(5 < X < 8) = P(-0.93 < z < 1.07) = 0.8577 - 0.1762 = 0.6815$

• Top 5% of wait times: $z_{0.95} = 1.645$

• $x = 6.4 + 1.645 \times 1.5 = 8.87$ min

Sampling Distributions

Sampling Distribution of the Mean

The sampling distribution of the sample mean describes the distribution of sample means from all possible samples of a given size from a population.

• Mean: $\mu_{\bar{x}} = \mu$

• Standard Deviation (Standard Error): $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

• Shape: If the population is normal or $n \geq 30$, the sampling distribution is approximately normal (Central Limit Theorem).

Sampling Distribution of a Proportion

• Mean: $\mu_{\hat{p}} = p$

• Standard Deviation: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

Example: Energy Needs in Pregnancy

• Population mean kcal/day, $\sigma = 50$ kcal/day.

• Sample size $n = 20$.

• Standard error: $\sigma_{\bar{x}} = \frac{50}{\sqrt{20}} \approx 11.18$

• Probability a randomly selected woman has energy need over 2625 kcal/day:

  • $z = \frac{2625 - 2600}{50} = 0.5$

  • $P(X > 2625) = 1 - 0.6915 = 0.3085$

• Probability sample mean of 20 women exceeds 2625 kcal/day:

  • $z = \frac{2625 - 2600}{11.18} = 2.24$

  • $P(\bar{X} > 2625) = 1 - 0.9875 = 0.0125$ (unusual, since $< 0.05$)

Summary Table: Key Formulas

Concept	Formula	Description
Binomial Mean	$\mu_X = np$	Expected number of successes
Binomial Std. Dev.	$\sigma_X = \sqrt{np(1-p)}$	Spread of binomial distribution
Z-score	$z = \frac{x - \mu}{\sigma}$	Standardized value
Sampling Mean	$\mu_{\bar{x}} = \mu$	Mean of sample means
Sampling Std. Error	$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$	Standard deviation of sample means
Sampling Proportion Mean	$\mu_{\hat{p}} = p$	Mean of sample proportions
Sampling Proportion Std. Dev.	$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$	Std. dev. of sample proportions

Additional info:

• Continuity correction is often used when approximating binomial probabilities with the normal distribution, by adjusting the discrete x-value by 0.5.

• Normal probability plots are a graphical tool for assessing normality; strong linearity suggests normality.