15 Problems Solved
Tutorial 4 focuses on the two most important continuous distributions in engineering: the Normal Distribution (Gaussian) and the Exponential Distribution. Master the use of Z-tables and the memoryless propertyβthey are guaranteed exam topics.
A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year. Assuming normal distribution, find the probability that a given battery will last less than 2.3 years.
To find probabilities for any normal distribution \(N(\mu, \sigma^2)\), we convert it to the Standard Normal Distribution \(Z \sim N(0, 1)\).
\(\mu = 3.0\), \(\sigma = 0.5\), \(X = 2.3\)
We need \(P(X < 2.3)=P(Z < -1.4)\).
Using symmetry: \(P(Z < -1.4)=1 - \Phi(1.4)\).
From Z-table, \(\Phi(1.4) \approx 0.9192\).
An electrical firm manufactures light bulbs with mean life 800 hours and SD 40 hours (normally distributed). Find the probability that a bulb burns between 778 and 834 hours.
\(Z_1 = \frac{778 - 800}{40} = -0.55\)
\(Z_2 = \frac{834 - 800}{40} = 0.85\)
\(P(-0.55 < Z < 0.85)=\Phi(0.85) - \Phi(-0.55)\)
\(= \Phi(0.85) - [1 - \Phi(0.55)]\)
\(\approx 0.8023 - [1 - 0.7088] = 0.8023 - 0.2912 = 0.5111\)
Diameter of a ball bearing follows \(N(3.0, 0.005^2)\). Specs are \(3.0 \pm 0.01\) cm. On average, how many ball bearings will be scrapped?
Scrapped means outside the spec limits. Find the probability of being inside the spec first, then subtract from 1.
\(P(\text{Scrapped}) = 1 - P(\text{Within Specs})\)
Lower Limit = 2.99 cm, Upper Limit = 3.01 cm.
\(P(2.99 < X < 3.01)=P\left(\frac{2.99-3.0}{0.005} < Z < \frac{3.01-3.0}{0.005}\right)\)
\(= P(-2 < Z < 2)=\Phi(2) - \Phi(-2)\)
\(= 0.9772 - 0.0228 = 0.9544\)
Scrap Rate \(= 1 - 0.9544 = 0.0456\).
Gauges reject components not within \(1.50 \pm d\). Measurement is \(N(1.50, 0.2^2)\). Determine \(d\) such that the specifications cover 95% of the measurements.
95% coverage means the area between \(-z\) and \(+z\) is 0.95. This relates to the critical value in statistics.
Instead of finding probability from Z, we find Z from probability.
For central 95% area, we have 2.5% in each tail.
\(\Phi(z) = 0.975 \implies z \approx 1.96\).
\(Z = \frac{\text{Absolute Deviation}}{\sigma} = \frac{d}{0.2}\)
\(1.96 = \frac{d}{0.2} \implies d = 0.392\).
Resistors have mean 40 ohms and SD 2 ohms (Normal). What percentage of resistors will have resistance exceeding 43 ohms?
Find the Z-score for 43 and then find the area to the right.
\(Z = \frac{43 - 40}{2} = 1.5\)
\(P(X > 43) = P(Z > 1.5) = 1 - \Phi(1.5)\)
\(= 1 - 0.9332 = 0.0668\)
Find the percentage of resistances exceeding 43 ohms if resistance is measured to the nearest ohm.
Rounding to 43 means the true value is between 42.5 and 43.5. Exceeding 43 means being 44 or higher.
To account for rounding, we look at the boundary halfway between the rounding points. To exceed 43, the value must be \( \ge 43.5 \).
To exceed 43 (when rounded to nearest integer), the true value \(X\) must be \(> 43.5\).
\(Z = \frac{43.5 - 40}{2} = 1.75\)
\(P(X > 43.5) = 1 - \Phi(1.75) = 1 - 0.9599 = 0.0401\)
Average grade 74, SD 7. If top 12% get A, what is the lowest possible A and highest possible B? (Normal curve assumption)
Top 12% means finding the Z-value where 88% of the distribution is below it.
From Z-table, \(z \approx 1.175\).
\(X = \mu + z\sigma\)
\(X = 74 + (1.175 \times 7) = 74 + 8.225 = 82.225\)
Using the exam data (\(\mu=74, \sigma=7\)), find the sixth decile.
Sixth decile (\(D_6\)) is the 60th percentile.
Points that split data into 10% increments. \(D_6 = \Phi(z) = 0.60\).
For cumulative area = 0.60, \(z \approx 0.253\).
\(X = 74 + (0.253 \times 7) = 74 + 1.77 = 75.77\)
Time spent in a bank is exponentially distributed with mean 10 minutes. (a) P(time > 15), (b) P(time > 15 | time > 10).
Parameter \(\lambda = 1/10 = 0.1\). Use the memoryless property for part (b).
Past duration doesn't affect future likelihood: \(P(X > s+t \mid X > s) = P(X > t)\).
\(P(X > 15) = e^{-0.1 \times 15} = e^{-1.5} \approx 0.2231\)
\(P(X > 15 \mid X > 10) = P(X > 5) \text{ (due to memoryless poperty)}\)
\(= e^{-0.1 \times 5} = e^{-0.5} \approx 0.6065\)
Jones and Brown are being served. Smith enters and starts service as soon as one leaves. If service times are exponential, what is the probability Smith is the last to leave?
No calculation needed. Think about the state of the system the moment Smith starts his service. Who is left?
Symmetry and Memoryless property of the Exponential distribution.
1. Smith starts his service when either Jones or Brown finishes. Let's say Jones leaves.
2. At this instant, Brown is still in service. Due to memoryless property, Brown's remaining time has the exact same distribution as a fresh service time.
3. Smith is also starting a fresh service time. Both are identical independent exponential variables.
4. By symmetry, the probability that Smith finishes after Brown is 1/2.
Repair time T is exponential with mean 0.5 hours. (a) P(T > 0.5), (b) P(T β₯ 12.5 | T > 12).
\(\lambda = 1/0.5 = 2\).
Memoryless property transforms the conditional probability into a simple one.
\(P(T > 0.5) = e^{-2 \times 0.5} = e^{-1} \approx 0.3679\)
\(P(T \ge 12.5 \mid T > 12) = P(T \ge 0.5)\)
\(= e^{-2 \times 0.5} = e^{-1} \approx 0.3679\)
Lifetime of a radio is exponential with mean 10 years. If you buy a 10-year-old radio, what's the probability it works another 10 years?
This is the counter-intuitive part of exponential distribution: things don't "wear out."
Probability given survival for 10 years, to survive another 10:
\(P(X > 20 \mid X > 10) = P(X > 10)\)
\(= e^{-(1/10) \times 10} = e^{-1} \approx 0.3679\)
Two games. Points: Win=2, Tie=1, Loss=0. Game 1: P(Not lose)=0.4. Game 2: P(Not lose)=0.7. Not losing implies Win/Tie equally likely. Find PMF of total points.
List possible outcomes for each game: {0, 1, 2}. Combine them using multiplication rule.
PMF of sum of independent variables.
P(0)=0.6, P(1)=0.2, P(2)=0.2
P(0)=0.3, P(1)=0.35, P(2)=0.35
1st win wins match, else match is drawn after 10 draws. G probabilities: Fischer win 0.4, Spassky win 0.3, Draw 0.3. Find Fischer's win probability and match duration PMF.
Fischer wins if he wins in game 1, or draws game 1 and wins game 2, etc.
Sum of a finite geometric series.
\( \sum_{k=1}^{10} (0.3)^{k-1} \times 0.4 = 0.4 \frac{1-0.3^{10}}{1-0.3} \approx 0.5714 \)
\( P(T=k) = (0.3)^{k-1}(0.7) \) for \( k \in [1, 9] \)
\( P(T=10) = 0.3^9 \)
Compute probabilities for events \(\{X > k\sigma\}\) and \(\{|X| > k\sigma\}\) for \(k=1,2,3\).
These values are the "magic numbers" of statistics. You'll recognize the 95% and 99.7% rules here.
| k | P(X > kΟ) | P(|X| > kΟ) |
|---|---|---|
| 1 | 0.1587 | 0.3174 |
| 2 | 0.0228 | 0.0456 |
| 3 | 0.0013 | 0.0026 |