PSTAT 5A Practice Worksheet 5 - SOLUTIONS

Continuous Random Variables and Confidence Intervals

Section A Solutions: Continuous Random Variables

Solution. Solution A1: Distribution Identification and Properties

(a) Exponential Distribution

Since the average time between arrivals is \(2\) minutes, we have:

• Parameter \(\lambda\): The rate parameter \(\lambda = \frac{1}{\mu} = \frac{1}{2} = 0.5\) arrivals per minute

• Probability calculation: \(P(X ≤ 1)\) where \(X \sim Exponential(0.5)\)

  For exponential distribution: \(P(X ≤ x) = 1 - e^{(- \lambda x)}\)

  \(P(X ≤ 1) = 1 - e^{(-0.5×1)} = 1 - e^{(-0.5)} = 1 - 0.6065 = \boxed{0.3935}\)

(b) Uniform Distribution

• Parameters: \(a = 10, b = 30\)

• Expected Value: \(E[X] = (a + b)/2 = \frac{(10 + 30)}{2} = \boxed{20}\)

• Variance: \(Var(X) = \frac{(b - a)^2}{12} = \frac{(30 - 10)^2}{12} = \frac{400}{12} = \boxed{33.3333}\)

Solution. Solution A2: Normal Distribution Calculations

Given: \(X \sim N(64, 2.5^2)\)

(a) \(P(X > 67)\)

Step 1: Standardize

\(Z = (67 - 64)/2.5 = 3/2.5 = 1.2\)

Step 2: Find probability

\(P(X > 67) = P(Z > 1.2) = 1 - P(Z ≤ 1.2) = 1 - 0.8849 = \boxed{0.1151}\)

(b) 25th percentile

Step 1: Find \(z\)-value for \(25\) -th percentile

\(P(Z ≤ z) = 0.25\), so \(z_{0.25} = -0.6745\)

Step 2: Convert back to \(X\)

\(x = \mu + z \sigma = 64 + (-0.6745)(2.5) = 64 - 1.6863 = \boxed{62.3137} \quad \text{inches}\)

(c) P(62 < X < 68)

Step 1: Standardize both values

\(Z_1 = (62 - 64)/2.5 = -0.8\) \(Z_2 = (68 - 64)/2.5 = 1.6\)

Step 2: Find probability

\(P(62 < X < 68) = P(-0.8 < Z < 1.6) = P(Z < 1.6) - P(Z < -0.8)\)

\(= 0.9452 - 0.2119 = \boxed{0.7333}\)

Section B Solutions: Confidence Intervals

Solution. Solution B1: Understanding Confidence Intervals

(a) Explanation of \(95\%\) Confidence Interval:

A \(95\%\) confidence interval means that if we were to repeat our sampling process many times (say \(100\) times) and construct a confidence interval each time using the same method, approximately \(95\) of those intervals would contain the true population mean. It does NOT mean there’s a \(95\%\) probability that the population mean lies in any one specific interval.

(b) Sample mean and margin of error:

Given \(CI\): (\(150g, 170g\))

• Sample mean: \(\bar x = (150 + 170)/2 = \boxed{160g}\)

• Margin of error: \(E = (170 - 150)/2 = \boxed{10g}\)

(c) True or False statement:

FALSE. Once we calculate a specific confidence interval, the population mean either is or isn’t in that interval, there’s no probability involved for that specific interval. The \(95\%\) refers to the long-run success rate of the method, not the probability for any individual interval.

Solution. Solution B2: Constructing Confidence Intervals

Given: \(n = 36, \bar x = 78.5, s = 12\)

(a) 95% Confidence Interval:

Step 1: Check conditions

• \(n = 36 ≥ 30\), so we can use \(z\)-distribution
• For \(95% CI: z{0.025} = 1.96\)

Step 2: Calculate margin of error

\(E = z_{0.025} × (\frac{s}{\sqrt{n}}) = 1.96 × (\frac{12}{\sqrt{36}}) = 1.96 × (\frac{12}{6}) = 1.96 × 2 = 3.92\)

Step 3: Construct interval

\(CI = \bar x ± E = 78.5 ± 3.92 = \boxed{(74.58, 82.42)}\)

(b) Interpretation:

We are \(95\%\) confident that the true population mean test score is between \(74.58\) and \(82.42\) points.

(c) Effects on interval width:

• Increasing confidence level to 99%: The interval would become wider because we need \(z_{0.005} = 2.576 > 1.96\)

• Increasing sample size to 144: The interval would become narrower because the margin of error would be \(E = 1.96 × (\frac{12}{\sqrt{144}}) = 1.96 × 1 = 1.96\) (smaller than \(3.92\))

Solution. Solution B3: Sample Size Determination

Given: \(E = \$5\), confidence = \(95\%, \sigma = \$25\)

(a) Required sample size:

Step 1: Use sample size formula

\(n = (z_{0.025} × \frac{\sigma}{E})^2\)

Step 2: Substitute values

\(n = (1.96 × 25 / 5)^2 = (49/5)^2 = 9.8^2 = 96.04\)

Step 3: Round up

\(\boxed{n = 97}\) customers (always round up for sample size)

(b) For margin of error = $3:

\(n = (1.96 × 25 / 3)^2 = (49/3)^2 = 16.333^2 = 266.67\)

\(\boxed{n = 267}\) customers

Optional Problem Solutions

Solution. Optional Solution 1: Conceptual Understanding

(a) Differences between discrete and continuous:

Values they can take:

• Discrete: Countable values (integers, specific points)

• Continuous: Uncountably infinite values (any real number in an interval)

How we calculate probabilities:

• Discrete: \(P(X = x)\) can be non-zero; we sum probabilities

• Continuous: \(P(X = x) = 0\) for any specific \(x\); we integrate over intervals

(b) Why \(P(X = x) = 0\) for continuous distributions:

In continuous distributions, there are infinitely many possible values in any interval. The probability of hitting any one exact value is infinitesimally small, hence zero. We instead calculate \(P(a < X < b)\) by integrating the PDF over the interval \([a,b]\).

(c) Relationship between PDF and CDF:

• PDF (f(x)): The probability density function gives the “density” of probability at each point

• CDF (F(x)): The cumulative distribution function gives \(P(X ≤ x)\)

• Relationship: \(F(x) = \int_{-∞}^x f(t)dt,\quad \text{and} \quad f(x) = F'(x)\)

Key Takeaways:

1. Always standardize normal distribution problems using \(Z = \frac{(X - μ)}{\sigma}\)

2. Interpret confidence intervals in context, they’re about the method’s reliability, not individual interval probabilities

3. Choose the right distribution use \(t\) when \(\sigma\) is unknown and \(n < 30\)

4. Round up sample sizes to ensure you meet the margin of error requirement

5. For continuous distributions, focus on intervals, not individual points