PSTAT 5A Practice Worksheet 5
Continuous Random Variables and Confidence Intervals
Instructions and Overview
⏰ Time Allocation:
Intro & Setup : 10 minutes
Section A (Continuous Distributions): 20 minutes
Section B (Confidence Intervals): 20 minutes
Optional Questions: Do on your own
Total: 50 minutes
📝 Important Instructions:
📚 Key Formulas Reference:
Continuous Random Variables:
Normal Distribution: \(X \sim N(\mu, \sigma^2)\)
- PDF: \(f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\)
- Standardization: \(Z = \frac{X - \mu}{\sigma}\) where \(Z \sim N(0,1)\)
- Mean: \(E[X] = \mu\)
- Variance: \(\text{Var}(X) = \sigma^2\)
Uniform Distribution: \(X \sim \text{Uniform}(a,b)\)
- PDF: \(f(x) = \frac{1}{b-a}\) for \(a \leq x \leq b\)
- Mean: \(E[X] = \frac{a+b}{2}\)
- Variance: \(\text{Var}(X) = \frac{(b-a)^2}{12}\)
Exponential Distribution: \(X \sim \text{Exponential}(\lambda)\)
- PDF: \(f(x) = \lambda e^{-\lambda x}\) for \(x \geq 0\)
- Mean: \(E[X] = \frac{1}{\lambda}\)
- Variance: \(\text{Var}(X) = \frac{1}{\lambda^2}\)
Confidence Intervals:
For Population Mean (σ known): \(\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\)
For Population Mean (σ unknown): \(\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\)
Margin of Error: \(E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\) or \(E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\)
Sample Size: \(n = \left(\frac{z_{\alpha/2} \cdot \sigma}{E}\right)^2\)
Section A: Continuous Random Variables
⏱️ Estimated time: 20 minutes
Problem A1: Distribution Identification and Properties
For each scenario below, identify the appropriate continuous distribution and find the requested values:
(a) The time (in minutes) between arrivals at a coffee shop follows an exponential distribution with an average of 2 minutes between arrivals.
- What is the parameter \(\lambda\)?
- What is the probability that the next customer arrives within 1 minute?
(b) A random number generator produces values uniformly between 10 and 30.
- What are the parameters a and b?
- What is the expected value and variance?
Work Space:
Problem A2: Normal Distribution Calculations
The heights of adult women in the US are normally distributed with \(\mu = 64\) inches and \(\sigma = 2.5\) inches.
(a) What is the probability that a randomly selected woman is taller than \(67\) inches?
(b) What height represents the \(25\)th percentile?
(c) What is the probability that a randomly selected woman has a height between \(62\) and \(68\) inches?
Remember to standardize: Convert to \(Z\)-scores using \(Z = \frac{X - \mu}{\sigma}\)
For part (b), you’re looking for the value \(x\) such that \(P(X ≤ x) = 0.25\)
Work Space:
Section B: Confidence Intervals
⏱️ Estimated time: 20 minutes
Problem B1: Understanding Confidence Intervals
(a) Explain in your own words what a \(95\%\) confidence interval means.
(b) A \(90\%\) confidence interval for the mean weight of apples is (150g, 170g). What is the sample mean and margin of error?
(c) True or False: “There is a \(95\%\) probability that the population mean lies within our calculated \(95\%\) confidence interval.” Explain your reasoning.
Work Space:
Problem B2: Constructing Confidence Intervals
A sample of \(36\) students has a mean test score of \(78.5\) with a standard deviation of \(12\).
(a) Construct a \(95\%\) confidence interval for the population mean test score.
(b) Interpret this interval in the context of the problem.
(c) What would happen to the width of the interval if:
We increased the confidence level to \(99\%\)?
We increased the sample size to \(144\)?
Decision Guide:
Use \(z\)-distribution when \(\sigma\) is known OR \(n ≥ 30\)
Use \(t\)-distribution when \(\sigma\) is unknown AND \(n < 30\)
For \(95\%\) CI: \(z_{0.025} = 1.96\)
Work Space:
Optional Questions
Optional Problem: Conceptual Understanding
(a) Explain the key difference between discrete and continuous random variables in terms of:
The values they can take
How we calculate probabilities
(b) Why do we use \(P(X = x) = 0\) for any specific value \(x\) in a continuous distribution?
(c) What’s the relationship between PDF and CDF for continuous distributions?
Work Space:
📋 Quick Reference:
Common Z-values:
\(90\%\) CI: \(z_{0.05} = 1.645\)
\(95\%\) CI: \(z_{0.025}\) = 1.96$
\(99\%\) CI: \(z_{0.005}\) = 2.576$
Common t-values (selected):
\(df = 24, \alpha = 0.05: t_{0.025} = 2.064\)
\(df = 35, \alpha = 0.05: t_{0.025} = 2.030\)