Interactive Guided Walkthrough
The probability density function of \(X\), the lifetime of a certain type of electronic device (measured in hours), is given by:
Drag the slider to change the threshold \(x_0\).
Calculated Probabilities:
\(P(X > x_0) = \int_{x_0}^{\infty} \frac{10}{x^2} dx = \) 0.500
\(F(x_0) = P(X \le x_0) = 1 - \frac{10}{x_0} = \) 0.500
(a) Find \(P(X > 20)\):
$$ P(X > 20) = \int_{20}^{\infty} \frac{10}{x^2} \, dx = \left[ -\frac{10}{x} \right]_{20}^{\infty} = 0 - (-\frac{10}{20}) = \frac{1}{2} $$
(b) CDF \(F(x)\):
For \(x > 10\): $$ F(x) = \int_{10}^{x} \frac{10}{t^2} \, dt = \left[ -\frac{10}{t} \right]_{10}^{x} = 1 - \frac{10}{x} $$
Consider 6 devices. We want to find the probability that at least 3 function for at least 15 hours.
Step 1: Probability for a single device
Threshold: 15 hours
\(p = P(X \ge 15) = 1 - F(15) = 1 - (1 - \frac{10}{15}) = \frac{2}{3} \approx 0.667\)
Step 2: Binomial Distribution \(Y \sim \text{Bin}(6, 2/3)\)
Find \(P(Y \ge 3)\)
Total Probability \(P(Y \ge 3) = \) 0.8999
(Sum of bars for k=3, 4, 5, 6)
$$ P(Y \ge 3) = \sum_{k=3}^{6} \binom{6}{k} (2/3)^k (1/3)^{6-k} $$
Calculating terms:
Sum \(\approx 0.8999\)
Compute \(E[X]\) for the following density functions.
$$ f(x) = \frac{3}{4}(1-x^2), \quad -1 < x < 1 $$
The expectation is \( E[X] = \int_{-1}^{1} x f(x) \, dx \).
Observation:
The function \(f(x)\) (blue) is even (symmetric about y-axis).
The integrand \(x f(x)\) (red) is odd (symmetric about origin).
Result: \( E[X] = 0 \)
$$ f(x) = \frac{5}{x^2}, \quad x > 5 $$
The expectation is \( E[X] = \int_{5}^{\infty} x \cdot \frac{5}{x^2} \, dx = \int_{5}^{\infty} \frac{5}{x} \, dx \).
Value of Integral \(\int_{5}^{M} \frac{5}{x} dx\):
14.98
As \(M \to \infty\), the area grows logarithmically without bound.
Result: \( E[X] = \infty \)
The density function of \(X\) is given by \( f(x) = a + bx^2 \) for \( 0 \le x \le 1 \).
Given that \( E[X] = 3/5 \), find \(a\) and \(b\).
Adjust \(a\) and \(b\) to satisfy the conditions.
System of Equations:
1. Normalization: \(\int_0^1 (a+bx^2)dx = a + b/3 = 1 \implies 3a + b = 3\)
2. Expectation: \(\int_0^1 x(a+bx^2)dx = a/2 + b/4 = 3/5 \implies 10a + 5b = 12\)
Solution:
\(a = 0.6 = 3/5\)
\(b = 1.2 = 6/5\)
If \(X \sim \text{Uniform}(0, 1)\), find the PDF of \(Y = e^X\).
Generate random samples from \(X\) and transform them to \(Y\).
Theoretical PDF of Y:
$$ f_Y(y) = \frac{1}{y}, \quad 1 < y < e $$
Transformation Method:
\(y = e^x \implies x = \ln y\)
\(dx/dy = 1/y\)
\(f_Y(y) = f_X(x(y)) |dx/dy| = 1 \cdot (1/y) = 1/y\)
Range: \(0 < x < 1 \implies e^0 < y < e^1 \implies 1 < y < e\)
You arrive at a bus stop at 10 a.m., knowing that the bus will arrive at some time uniformly distributed between 10 and 10:30.
(a) What is the probability that you will have to wait longer than 10 minutes?
(b) If, at 10:15, the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes?
Probability that you wait longer than 10 minutes.
Total Interval: [0, 30]
Target Interval: (10, 30]
Probability = \(\frac{30-10}{30} = \frac{20}{30} = \frac{2}{3}\)
Given bus hasn't arrived by 10:15, prob waiting > 10 more mins?
Condition: \(X > 15\) (Interval length 15)
Target: \(X > 15 + 10 = 25\) (Interval length 5)
Probability = \(\frac{5}{15} = \frac{1}{3}\)
(a) \(X \sim U(0, 30)\). \(P(X > 10) = \frac{30-10}{30-0} = \frac{20}{30} = \frac{2}{3}\).
(b) We want \(P(X > 25 | X > 15)\).
$$ P(X > 25 | X > 15) = \frac{P(X > 25 \cap X > 15)}{P(X > 15)} = \frac{P(X > 25)}{P(X > 15)} $$
$$ = \frac{(30-25)/30}{(30-15)/30} = \frac{5/30}{15/30} = \frac{5}{15} = \frac{1}{3} $$
If \(X\) is a normal random variable with parameters \(\mu = 10\) and \(\sigma^2 = 36\), compute:
Select Problem Part:
Calculated Probability:
0.0000
Select a problem part...
Standardization: \(Z = \frac{X - \mu}{\sigma} = \frac{X - 10}{6}\)
The salaries of physicians in a certain area are approximately normally distributed. If 25 percent of these physicians earn less than $180,000 and 25 percent earn more than $320,000, approximately what fraction earn:
Adjust Mean (\(\mu\)) and Std Dev (\(\sigma\)) to match the conditions.
(a) \(P(X < 200) = \)
(b) \(P(280 < X < 320) = \)
Symmetry: Tails are equal (25%), so \(\mu\) is midpoint of 180 and 320.
\(\mu = (180+320)/2 = 250\).
Sigma: \(P(X < 320) = 0.75 \implies Z_{0.75} \approx 0.675\).
\(0.675 = (320-250)/\sigma \implies \sigma \approx 103.7\).
(a) \(P(X < 200) \approx 0.3156\)
(b) \(P(280 < X < 320) \approx 0.1359\)
If 65 percent of the population of a large community is in favor of a proposed rise in school taxes, approximate the probability that a random sample of 100 people will contain:
Use the central limit theorem.
Parameters:
\(n = 100, p = 0.65\)
\(\mu = np = 65\)
\(\sigma = \sqrt{np(1-p)} \approx 4.77\)
Approximation:
Select a case
(a) \(X \ge 50\): \(P(Y > 49.5) \approx 1\)
(b) \(60 \le X \le 70\): \(P(59.5 < Y < 70.5) \approx 0.7498\)
(c) \(X < 75\): \(P(Y < 74.5) \approx 0.9767\)
One thousand independent rolls of a fair die will be made. Compute an approximation to the probability that the number 6 will appear between 150 and 200 times inclusively. If the number 6 appears exactly 200 times, find the probability that the number 5 will appear less than 150 times. Use the central limit theorem.
Approximate probability that the number 6 appears between 150 and 200 times.
$$ n=1000, p=1/6 \implies \mu \approx 166.67, \sigma \approx 11.79 $$
Calculation (with continuity correction):
$$ P(149.5 < Y < 200.5) $$
Z-scores: \( z_1 \approx -1.46, z_2 \approx 2.87 \)
Result: 0.9258
Given exactly 200 sixes, prob that 5 appears < 150 times.
Remaining rolls: \( 1000 - 200 = 800 \). Outcomes: {1,2,3,4,5}.
$$ P(5|\text{not } 6) = 0.2 \implies \mu' = 160, \sigma' \approx 11.31 $$
Calculation:
$$ P(Y' < 149.5) $$
Z-score: \( z \approx -0.93 \)
Result: 0.1762
The time (in hours) required to repair a machine is an exponentially distributed random variable with parameter \(\lambda = 1/2\). What is
Probability repair takes longer than 2 hours.
Calculation:
$$ P(X > 2) = e^{-0.5(2)} = e^{-1} $$
Result: \(\approx 0.3679\)
\( P(X \ge 10 \mid X > 9) \)
Given it has already taken 9 hours, what is the prob it takes at least 1 more hour?
Calculation:
$$ P(X \ge 10 \mid X > 9) = P(X \ge 1) $$
$$ = e^{-0.5(1)} = e^{-0.5} $$
Result: \(\approx 0.6065\)
The number of years a radio functions is exponentially distributed with parameter \(\lambda = 1/8\).
If Jones buys “a 2-years old radio”, what is the probability that it will be working after an additional 8 years?
Explore how the conditional probability depends only on the additional time.
Calculations:
Total required life: \(s + t = \) 10 years
\(P(X > s+t \mid X > s) = \frac{P(X > s+t)}{P(X > s)} = \) 0.3679
\(P(X > t) = e^{-\lambda t} = \) 0.3679
Notice they are always equal!
Solution:
\(X \sim \text{Exponential}(\lambda = 1/8)\).
We want \(P(X > 2 + 8 \mid X > 2) = P(X > 10 \mid X > 2)\).
By the Memoryless Property:
$$ P(X > s + t \mid X > s) = P(X > t) $$
Here \(s=2, t=8\). So we just need \(P(X > 8)\).
$$ P(X > 8) = e^{-(1/8) \cdot 8} = e^{-1} \approx 0.3679 $$