Statistics I · Unit 6 · 12 hrs
Probability Distributions
Exam-focused notes for Probability Distributions (Statistics I, STA169): what the TU syllabus asks and how it has actually been tested, with 24 solved past questions from this unit.
What this unit covers
- Probability distribution function
- Joint probability distribution of two random variables
- Discrete distributions: Bernoulli trial, Binomial and Poisson distributions
- Continuous distribution: Normal distributions
- Standardization of normal distribution
- Normal distribution as an approximation of Binomial and Poisson distribution
- Exponential, Gamma distribution
- Problems and illustrative examples related to computer Science and IT
Joint probability distribution of two random variables
The joint density function of two continuous random variables X and Y is f(x, y) = kxy 0 < x < 4, 1 < y < 5= 0 otherwisea. Find the value of constant k. b. Find P(x > 3, y < 2) c. Find P(1 < x < 2, 2 < y < 3)[10]
$$f(x, y) = kxy, \quad 0 < x < 4, \quad 1 < y < 5$$ $$f(x, y) = 0, \quad \text{otherwise}$$ Required: - (a) constant $k$ - (b) $P(X 3, Y < 2)$ - (c) $P(1 < X < 2, 2 < Y < 3)$ --- Normalization condition: $$\int{1}^{5} \int{0}^{4} kxy \, dx \, dy = 1$$ Inner...
Full solved answer →Fatigue Test Probability Analysis
The following joint probability data apply to a fatigue test run on bronze strips. X represents cycles to failure (in $10^5$) when alternate strips are bent at a high level of deflection. Y represents the same at a lower deflection event.
$$\begin{array}{c|cccc} X/Y & 20 & 30 & 40 & 50 \ \hline 4 & 0.01 & 0.03 & 0.05 & 0.02 \ 5 & 0.03 & 0.10 & 0.08 & 0.04 \ 6 & 0.02 & 0.08 & 0.12 & 0.11 \ 7 & 0.02 & 0.04 & 0.07 & 0.18 \ \end{array}$$
a. Find the marginal probability distribution for X and Y
b. Determine the conditional probability distribution of Y given $X = 5$
c. Are X and Y independent?
[5]
Joint probability table $f(x,y)$: X/Y 20 30 40 50 ------------------------- 4 0.01 0.03 0.05 0.02 5 0.03 0.10 0.08 0.04 6 0.02 0.08 0.12 0.11 7 0.02 0.04 0.07 0.18 Marginal of X: $P(X=x)=\sumy f(x,y)$ $$P(X=4)=0.01+0.03+0.05+0.02=0.11$$ $$P(X=5)=0.03+0.10+0...
Full solved answer →If two random variables have the joint probability density function find (i) constant k (ii) conditional probability density function of X (iii) Identify whether X and Y are independent.
$$f(x,y) = \begin{cases} k(2x + 3y), & 0 \leq x \leq 1, 0 \leq y \leq 1 \ 0, & \text{otherwise} \end{cases}$$
[5]
- $f(x,y) = k(2x+3y)$ for $0 \le x \le 1$, $0 \le y \le 1$; zero otherwise. --- Normalization condition: $$\int0^1 \int0^1 k(2x+3y)\,dx\,dy = 1$$ Inner integral over $x$: $$\int0^1 (2x+3y)\,dx = \left[x^2 + 3xy\right]0^1 = 1 + 3y$$ Outer integral over $y$: ...
Full solved answer →If two random variables have the joint probability density function $$f(x,y) = \begin{cases} k(2x + 3y), & 0 \leq x \leq 1, 0 \leq y \leq 1 \ 0, & \text{otherwise} \end{cases}$$
find (i) constant $k$ (ii) conditional probability density function of $X$ (iii) Identify whether $X$ and $Y$ are independent. [5]
- Joint PDF: $f(x,y) = k(2x+3y)$ for $0 \le x \le 1$, $0 \le y \le 1$; zero otherwise. --- The total probability must equal 1: $$\int0^1 \int0^1 k(2x+3y)\, dx\, dy = 1$$ Inner integral (over x): $$\int0^1 (2x+3y)\, dx = \left[x^2 + 3xy\right]0^1 = 1 + 3y$$ ...
Full solved answer →Random Variable and Bivariate Probability Distribution
Definition of a Random Variable: A random variable is a function that assigns numerical values to the outcomes of a random experiment. It can be discrete (taking countable values) or continuous (taking any value in an interval).
A random variable is a real-valued function that assigns a numerical value to each outcome in the sample space of a random experiment. A random variable is discrete if it takes finitely many or countably many values, each with a probability $P(X=x)$ (the pr...
Full solved answer →If two random variables have the joint probability density function
$$f(x,y) = \begin{cases} k e^{-(x+y)}, & 0 < x < \infty, 0 < y < \infty \ 0, & \text{otherwise} \end{cases}$$
Find (i) constant $k$ (ii) conditional probability density function of $X$ given $Y$ (iii) Var$(3X + 2Y)$. [5]
Joint PDF: $$f(x,y) = \begin{cases} k\,e^{-(x+y)}, & 0<x<\infty,\ 0<y<\infty \\ 0, & \text{otherwise} \end{cases}$$ Find: (i) constant $k$, (ii) conditional PDF of $X$ given $Y$, (iii) $\text{Var}(3X+2Y)$. --- $$\int0^\infty\int0^\infty k\,e^{-(x+y)}\,dx\,d...
Full solved answer →Let X and Y be two continuous random variable having joint pdf $f(x, y) = c(x^2 + y^2)$, $0 < x < 1$, $0 < y < 1$, $= 0$ otherwise. Determine (a) the value of c. (b) $P(x < 0.5, y > 0.5)$. [5]
$$f(x, y) = c(x^2 + y^2), \quad 0 < x < 1,\ 0 < y < 1$$ $$f(x, y) = 0 \quad \text{otherwise}$$ Required: (a) value of $c$; (b) $P(X < 0.5,\ Y 0.5)$. --- Normalization condition: $$\int{0}^{1}\int{0}^{1} c(x^2 + y^2)\, dx\, dy = 1$$ Inner integral (over x): ...
Full solved answer →Discrete distributions
Define binomial distribution. Under what conditions is the binomial distribution appropriate? The probability of a novice archer hitting the target with any shot is 0.3. Given that the archer shoots six arrows, find the probability that the target is hit at least once. [5]
- Probability of success (hit) per shot: $p = 0.3$ - Probability of failure (miss): $q = 1 - 0.3 = 0.7$ - Number of independent trials (arrows): $n = 6$ - Required: $P(\text{at least one hit}) = P(X \ge 1)$ All required data present. A binomial distribution...
Full solved answer →Question
Under what conditions does Binomial distribution tend to Poisson distribution? The number of telephone calls received during the month of May is summarized in the following table. Fit the Poisson distribution.
$$\begin{array}{c|ccccc} \text{Number of telephone calls per day} & 0 & 1 & 2 & 3 & 4 \ \hline \text{Number of days} & 8 & 12 & 18 & 13 & 9 \ \end{array}$$
[10]
Number of calls per day (x) 0 1 2 3 4 ------------------ Number of days (f) 8 12 18 13 9 $N = \sum f = 8 + 12 + 18 + 13 + 9 = 60$ --- The Binomial distribution $B(n, p)$ tends to the Poisson distribution when: 1. Number of trials is very large: $n \to \inft...
Full solved answer →The mean and variance of the number of flights arriving late in a day are 2 and 1.6 respectively. Assuming binomial distribution, are those values consistent? If yes, find the probability that (i) none of the flights are late today and (ii) at least one flight is late today. [5]
- Mean $= np = 2$ - Variance $= npq = 1.6$ For a binomial distribution, variance < mean always (since $q < 1$). Here mean $= 2$, variance $= 1.6$, and $1.6 < 2$, so the values are consistent. $$q = \frac{npq}{np} = \frac{1.6}{2} = 0.8$$ $$p = 1 - q = 0.2$$ ...
Full solved answer →a. What do you understand by Poisson distribution? What are its main features? b. What do you mean by joint probability distribution function? Write down its properties.[10]
--- Poisson distribution is a discrete probability distribution that is used to model the number of times an event occurs in a fixed interval of time or space, under the following conditions: - The probability of success is very small, i.e., p → 0 - The num...
Full solved answer →Fit a binomial distribution to the following data:
$$\begin{array}{c|ccccccc} X & 0 & 1 & 2 & 3 & 4 & 5 & 6 \ \hline f & 5 & 8 & 15 & 14 & 10 & 6 & 2 \ \end{array}$$
[5]
X 0 1 2 3 4 5 6 ------------------------ f 5 8 15 14 10 6 2 Maximum value of X is 6, so $n = 6$. X f fX ---------- 0 5 0 1 8 8 2 15 30 3 14 42 4 10 40 5 6 30 6 2 12 Total N = 60 ΣfX = 162 $$\bar{X} = \frac{\Sigma fX}{N} = \frac{162}{60} = 2.7$$ $$np = \bar{...
Full solved answer →A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 15%. The inspector randomly picks 10 items from a shipment. What is the probability that there will be at least one defective item among these 10? [5]
- Defective rate (probability of success): $p = 0.15$ - Non-defective probability: $q = 1 - p = 0.85$ - Sample size: $n = 10$ - Required: $P(X \geq 1)$ --- Each device is defective or not, selections are independent, and $p$ is constant. This is a Binomial ...
Full solved answer →Message arrives at an electronic message center at random times, with an average of 9 messages per hour. a. What is the probability of receiving at least four messages during the next hour? b. What is the probability of receiving at most three messages during the next hour? [5]
- Average rate: 9 messages per hour - Time interval: next 1 hour - Model: Poisson distribution - Parameter: $\lambda = 9$ Poisson probability mass function: $$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}, \quad \lambda = 9$$ With $e^{-9} = 0.00012341$. Using...
Full solved answer →What do you understand by binomial distribution? What are its main features? What do you mean by marginal probability distribution? Write down its properties.[10]
--- A binomial distribution is a discrete probability distribution that describes the number of successes in n independent trials, where each trial has only two possible outcomes: success (with probability p) and failure (with probability q = 1 - p). If a r...
Full solved answer →Write the properties of Poisson distribution. Fit a Poisson distribution and find the expected frequencies.
$$\begin{array}{c|cccccccc} \text{X} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \ \text{Y} & 71 & 112 & 117 & 57 & 27 & 11 & 3 & 1 \end{array}$$
[5]
Frequency distribution: X 0 1 2 3 4 5 6 7 --------------------------- Y (f) 71 112 117 57 27 11 3 1 All values readable. Nothing missing. --- Let $X \sim P(\lambda)$. 1. PMF: $P(X=x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!},\quad x=0,1,2,\dots$ 2. Mean = Varianc...
Full solved answer →Fitting a Binomial Distribution
Under what condition binomial probability distribution? Five unbiased coins are tossed 100 times and the following results were obtained. Fit the binomial distribution.
| No of heads | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Frequency | 5 | 24 | 35 | 22 | 10 | 4 |
[5]
The binomial probability distribution applies under the following conditions: - Each trial results in only two mutually exclusive outcomes (success/failure). - The number of trials n is fixed and finite. - The trials are independent of one another. - The pr...
Full solved answer →Define poisson probability distribution. Cars arrive at a petrol station at an average rate of 3 per minute. Assuming that the cars arrive at random, find the probability that (i) no car arrives during a particular minute. (ii) at least one car arrive during a particular minute. (iii) four cars arrives in any 2 minutes. [5]
The Poisson distribution is a discrete probability distribution that describes the number of events occurring in a fixed interval of time (or space), given that these events occur with a known constant average rate $\lambda$ and independently of each other....
Full solved answer →Continuous distribution
State features of normal distribution. In a photographic process, the developing time of prints as a random variable having normal distribution with mean of 18.25 seconds with standard deviation 0.34 seconds. Find the probability that at least 17.64 seconds to develop one of the prints. [5]
1. Bell-shaped and symmetric about the mean $\mu$; the two halves are mirror images. 2. Mean = Median = Mode, all located at the center $\mu$. 3. Completely defined by two parameters: mean $\mu$ and standard deviation $\sigma$. 4. Total area under the curve...
Full solved answer →What do you mean by normal distribution? From a batch of 10000, the lifetime of laptop batteries has a normal distribution with a mean of 40 months and a standard deviation of 8 months. What is the probability that a laptop selected at random will have life time (i) more than 50 months? (ii) between 40 and 50 months? [5]
Parameter Value ------------------ Batch size (N) 10,000 Mean (μ) 40 months Standard Deviation (σ) 8 months Required: (i) $P(X 50)$, (ii) $P(40 < X < 50)$ --- A normal distribution is a continuous probability distribution whose graph is a symmetric, bell-sh...
Full solved answer →The lifetime of a certain electronic component is a normal random variate with the expectation of 5000 hours and a standard deviation of 100 hours. Compute the probabilities under the following conditions:a. Lifetime of components between 3000 to 6500 hours b. Lifetime of components between 3000 to 6500 hours c. Lifetime of components more than 6000 hours [5]
- Mean $\mu = 5000$ hours - Standard deviation $\sigma = 100$ hours - $X \sim N(5000, 100^2)$ Standardization: $$Z = \frac{X - \mu}{\sigma} = \frac{X - 5000}{100}$$ Note on the question: Parts (a) and (b) are stated identically ("between 3000 to 6500 hours"...
Full solved answer →a. Define Normal distribution. What are the main characteristics of a Normal distribution? b. What do you mean by probability density function? Write down its properties.[10]
--- A Normal Distribution is a continuous probability distribution of a random variable X with parameters μ (mean) and σ² (variance). Its probability density function is given by: $$f(x) = \frac{1}{\sigma\sqrt{2\pi}} \cdot e^{-\frac{1}{2}\left(\frac{x-\mu}{...
Full solved answer →A certain machine makes electrical resistors having a mean resistance of 40 ohms and standard deviation of 2 ohms. Assuming that the resistance follows a normal distribution.(i) What percentage of resistors will have a resistance exceeding 43 ohms?(ii) What percentage of resistors will have a resistance between 30 ohms to 45 ohms? [5]
- Mean resistance: $\mu = 40$ ohms - Standard deviation: $\sigma = 2$ ohms - Distribution: Normal Standardization formula: $$Z = \frac{X - \mu}{\sigma}$$ --- Z-score for X = 43: $$Z = \frac{43 - 40}{2} = \frac{3}{2} = 1.5$$ Probability: $$P(X 43) = P(Z 1.5)...
Full solved answer →Define normal distribution. What are the main characteristics of normal distribution? Extruded plastic rods are automatically cut into length 5 inches. Actual length are normally distributed about a mean of 5 inches and their standard deviation is 0.05 inches. (i) What proportion of rods exceed tolerance limits of 4.9 inches to 5.1 inches? (ii) Proportion of rods having tolerance rod which is greater than 6.5 inches.[10]
A normal distribution is a continuous probability distribution, symmetrical and bell-shaped about its mean, defined by two parameters, mean $\mu$ and standard deviation $\sigma$. Its probability density function is: $$f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-...
Full solved answer →Make Unit 6 stick
Practice STA169 with flashcards & quizzes