Basic Statistics · Unit 6
Probability and Probability Distributions
Exam-focused notes for Probability and Probability Distributions (Basic Statistics, STA154): what the TU syllabus asks and how it has actually been tested, with 27 solved past questions from this unit.
What this unit covers
- Basic probability concepts and rules
- Conditional probability and Bayes theorem
- Random variables and probability functions
- Discrete probability distributions
- Binomial distribution
- Poisson distribution
- Continuous probability distributions
- Normal distribution properties and applications
- Standard normal distribution and Z-scores
Standard normal distribution and Z-scores
The distribution of monthly incomes of 5,000 employees of a certain industrial unit was found to be normally distributed with mean of Rs. 2,000 and a standard deviation of Rs. 200.(i) Estimate the range of incomes of the middle 60% employees.(ii) Estimate the lowest income of richest 10% employees.(iii) Estimate the highest income of poorest 10% employees.[10+0+0+0]
- Number of employees: $N = 5000$ - Distribution: Normal - Mean: $\mu = 2000$ - Standard deviation: $\sigma = 200$ Standardization: $Z = \dfrac{X - \mu}{\sigma}$, so $X = \mu + Z\sigma$. --- Middle 60% leaves 20% in each tail. - Lower cut-off: cumulative ar...
Full solved answer →The resolution time for customer support queries follows a normal distribution with mean 30 minutes and standard deviation 5 minutes. (a) What percentage of queries are resolved in less than 25 minutes? (b) What percentage are resolved between 25 and 35 minutes? μ=30,σ=5\mu = 30, \quad \sigma = 5μ=30,σ=5[5]
- Mean: $\mu = 30$ minutes - Standard deviation: $\sigma = 5$ minutes - Distribution: Normal, $X \sim N(30, 5^2)$ --- Standardize using $Z = \dfrac{X - \mu}{\sigma}$: $$Z = \frac{25 - 30}{5} = \frac{-5}{5} = -1$$ Look up the cumulative probability: $$P(Z < ...
Full solved answer →Define standard normal distribution. For a certain type of computers, the length of time between charges of the battery is normally distribbuted with a mean of 50 hours and a standard deviation of 15 hours. He owns one of these computers and wants to know the probability that the length of time will be (i) between 50 and 70 hours, (ii) more than 60 hours, and (iii) less than 45 hours?[10]
- Mean $\mu = 50$ hours - Standard deviation $\sigma = 15$ hours - $X \sim N(50, 15^2)$, normally distributed - Required: - (i) $P(50 \le X \le 70)$ - (ii) $P(X 60)$ - (iii) $P(X < 45)$ The standard normal distribution is a normal distribution with mean $\m...
Full solved answer →Basic probability concepts and rules
A piece of equipment will function only when all the components A, B, and C are working. The probability of A failing during one year is 0.15, that of B failing is 0.05, and that of C failing is 0.10. What is the probability that the equipment will not fail before the end of one year? [5]
- Equipment functions only when all of A, B, C are working. - $P(A \text{ fails}) = 0.15$ - $P(B \text{ fails}) = 0.05$ - $P(C \text{ fails}) = 0.10$ - Component failures assumed independent. - Find: $P(\text{equipment does not fail in one year})$ Probabili...
Full solved answer →The probability that an integrated circuit chip will have defective etching is 0.12, the probability that it will have crack defect is 0.29, and the probability that it has both defects is 0.07. What is the probability that a newly manufactured chip will have either an etching or crack defect. [5]
- P(Etching defect) = P(E) = 0.12 - P(Crack defect) = P(C) = 0.29 - P(Both defects) = P(E ∩ C) = 0.07 - Required: P(E ∪ C) Addition Rule of Probability: $$P(E \cup C) = P(E) + P(C) - P(E \cap C)$$ Substituting: $$P(E \cup C) = 0.12 + 0.29 - 0.07$$ $$P(E \cu...
Full solved answer →A new computer virus can enter the system through e-mail or through the internet. There is a 30% chance of receiving this virus hrough e-mail. There is a 40% chance of receiving it through the internet. Also, the virus enters the system simultaneously through e-mail and the internet with probability 0.15. What is, the probability that the virus does not enter the system at all? [5]
- $P(E)$ = probability virus enters through e-mail = $0.30$ - $P(I)$ = probability virus enters through internet = $0.40$ - $P(E \cap I)$ = probability virus enters through both simultaneously = $0.15$ Step 1: Probability that the virus enters the system Th...
Full solved answer →Suppose that after 10 years of service, 40% of computers have problems with motherboards (MB), 30% have problems with hard drives (HD), and l5% have problems with both MB and HD. What is the probability that a 10-year old computer still has fully functioning MB and HD? [5]
- $P(MB) = 0.40$ (probability of motherboard problem) - $P(HD) = 0.30$ (probability of hard drive problem) - $P(MB \cap HD) = 0.15$ (probability of both problems) - Find: $P(\text{no MB problem AND no HD problem})$ Step 1: Probability of at least one proble...
Full solved answer →The odds in against of A solving a problem as 8 to 6 and the odds in favor of B solving the same problem are 14 to 10. What is the probability that (a) both A and B will solve it? (b) A solve it but B fails to solve it? [5]
- Odds against A solving = 8 : 6 - Odds in favor of B solving = 14 : 10 For A (odds against = 8 : 6): Odds against means unfavorable : favorable = 8 : 6, so: $$P(A) = \frac{6}{8+6} = \frac{6}{14} = \frac{3}{7}$$ $$P(A') = 1 - \frac{3}{7} = \frac{4}{7}$$ For...
Full solved answer →Conditional probability and Bayes theorem
Three persons A, B, and C are being considered for appointment as Vice-Chancellor of a university, and whose chances of being selected are in the proportion 4:2:3 respectively. The probability that A, if selected, will introduce democratization is 0.3, and the corresponding probabilities for B and C are 0.5 and 0.8. What is the probability that democratization would be introduced? [5]
- Selection proportion for A : B : C = 4 : 2 : 3 - $P(D \mid A) = 0.3$ - $P(D \mid B) = 0.5$ - $P(D \mid C) = 0.8$ where $D$ = "democratization is introduced". Total parts $= 4 + 2 + 3 = 9$ $$P(A) = \frac{4}{9}, \quad P(B) = \frac{2}{9}, \quad P(C) = \frac{...
Full solved answer →A database server has a 95% reliability rate. Even if the server does not fail, there is a 5% chance that it will still report an error. If the server reports an error, what is the probability that the server has actually failed? $P(\text{Failure}) = 0.05, \quad P(\text{False positive}) = 0.05$ [5]
- Server reliability = 95%, hence probability of failure: $$P(F) = 0.05$$ - Probability of no failure: $$P(\bar{F}) = 1 - 0.05 = 0.95$$ - False positive rate (error reported even though no failure): $$P(E \mid \bar{F}) = 0.05$$ - Assumption from problem sta...
Full solved answer →Suppose Rajesh receives 50 messages and Harish receives 90 messages in the personal emails respectively. Please note that the email address of Rajesh and Harish is different. Rajesh receives 1% junk emails, and Harish receives 2% junk emails. A person is chosen at random at the end of a day and found the email message is junk. What is the probability that this junk email found in the Harish's email inbox? [5]
- Rajesh messages: 50, junk rate = 1% = 0.01 - Harish messages: 90, junk rate = 2% = 0.02 - Total messages: 50 + 90 = 140 - Find: $P(\text{Harish} \mid \text{Junk})$ Define events: - $R$ = email from Rajesh, $H$ = email from Harish, $J$ = email is junk Prio...
Full solved answer →Random variables and probability functions
A tech team records the number of bug reports closed per hour. The probability distribution is given below. Find the expected number of bug reports closed per hour and its variance.
$$\begin{array}{|c|ccccc|}\hline Y & 0 & 1 & 2 & 3 & 4 \ \hline P(Y) & 0.10 & 0.18 & 0.32 & 0.30 & 0.10 \ \hline \end{array}$$
[5]
Probability distribution of $Y$ (bug reports closed per hour): $Y$ 0 1 2 3 4 ------------------------------ $P(Y)$ 0.10 0.18 0.32 0.30 0.10 Check: $\sum P(Y) = 0.10 + 0.18 + 0.32 + 0.30 + 0.10 = 1.00$ ✓ (valid distribution) $$E(Y) = \sum Y \cdot P(Y)$$ $Y$ ...
Full solved answer →Define random variable. A random variable $X$ has following probability function. Find, (i) the value of $C$, (ii) $P(X < 3)$, $P(X \geq 1)$, (iii) mean and variance.
| X | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| P(X) | 0.1 | C | 0.2 | 2C | 0.3 | C |
[5]
A random variable is a real-valued function that assigns a numerical value to each outcome in the sample space of a random experiment. Random variables are classified as discrete (countable values) or continuous (values over an interval). X 0 1 2 3 4 5 ----...
Full solved answer →Define Random Variable and Solve
Define random variable: A random variable is a function that assigns numerical values to the outcomes of a random experiment. It can be discrete (taking specific values) or continuous (taking any value in an interval).
Given the probability distribution:
$$\begin{array}{|c|c|c|c|c|c|}\hline X & -2 & -1 & 0 & 1 & 2 \ \hline P(X) & 0.2 & K & 0.4 & 2K & K \ \hline \end{array}$$
A random variable is a real-valued function that assigns a numerical value to each outcome in the sample space of a random experiment. In other words, it maps the outcomes of a random phenomenon to numbers. Random variables are classified as: - Discrete ran...
Full solved answer →Probability Distribution of Computer Crashes
Following table represents the probability distribution for the number of computers crashes monthly in a reputed software company in Biratanagar. Compute mean and standard deviation of number of computer crashes and interpret them.
| Number of computer crashes | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Prob(X = x) | 0.10 | 0.20 | 0.45 | 0.15 | 0.05 | 0.05 |
[5]
Probability distribution: $x$ 0 1 2 3 4 5 --------------------- $P(X=x)$ 0.10 0.20 0.45 0.15 0.05 0.05 Check sum: $0.10+0.20+0.45+0.15+0.05+0.05 = 1.00 \checkmark$ $$\mu = E(X) = \sum x\,P(X=x)$$ $x$ $P(X=x)$ $xP(x)$ --------- 0 0.10 0.00 1 0.20 0.20 2 0.45...
Full solved answer →Binomial distribution
It is observed that 80% of television viewers watch an entertainment channel. What is the probability that at least 80% of the viewers in a random sample of five watch an entertainment channel? [5]
- Population proportion watching entertainment channel: $p = 0.80$ - Probability of not watching: $q = 1 - p = 0.20$ - Sample size: $n = 5$ - Required: $P(\text{at least } 80\% \text{ of sample watch})$ Distribution: Let $X$ = number of viewers (out of 5) w...
Full solved answer →Define binomial distribution. Fit a binomial distribution on the following data:
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| f | 28 | 62 | 46 | 10 | 4 |
[5]
$x$ 0 1 2 3 4 -------------------- $f$ 28 62 46 10 4 Maximum value of $x = 4 \Rightarrow n = 4$. A binomial distribution is a discrete probability distribution giving the number of successes in $n$ independent trials, each with two outcomes (success/failure...
Full solved answer →During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of chips received a thick enough coating, find the probability that among 15 chips (i) at least 12 will have thick enough coatings, (ii) at most 2 will have thick enough coatings, and (ii) exactly 10 will have thick enough coatings. [5]
- Probability a chip receives thick enough coating: $p = 0.70$ - Failure probability: $q = 1 - p = 0.30$ - Number of chips: $n = 15$ - $X$ = number of chips with thick enough coating $\sim \text{Binomial}(15, 0.70)$ Formula: $$P(X = k) = \binom{15}{k}(0.70)...
Full solved answer →During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of chips received a thick enough Coating. find the probability that among 15 chips (1) at least 12 will have thick enough coatings, and (2) exactly 10 will have thick enough coatings. [5]
- Probability of thick enough coating: $p = 0.70$ - Probability of not thick enough: $q = 1 - p = 0.30$ - Sample size: $n = 15$ Binomial distribution: $P(X = r) = \binom{n}{r} p^r q^{n-r}$ --- $$P(X \geq 12) = P(12) + P(13) + P(14) + P(15)$$ P(X = 12): $$\b...
Full solved answer →Poisson distribution
A server records the number of requests per minute, which follows a Poisson distribution with a mean of 5 requests per minute. (a) What is the probability that exactly 3 requests occur in a given minute? (b) What is the probability that more than 2 requests will occur? λ=5\lambda = 5λ=5[5]
- Distribution: Poisson - Mean rate: $\lambda = 5$ requests per minute - Interval: 1 minute - (a) Find $P(X = 3)$ - (b) Find $P(X 2)$ Poisson formula: $$P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}, \quad \lambda = 5$$ Note: $e^{-5} \approx 0.0067379$ --- $$...
Full solved answer →Fit a Poisson distribution to the following data.
| Defects (X = x) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Number of pages | 142 | 156 | 69 | 27 | 5 | 1 |
[5]
Defects (x) 0 1 2 3 4 5 --------------------- No. of pages (f) 142 156 69 27 5 1 $$\lambda = \bar{x} = \frac{\sum f x}{\sum f}$$ x f f·x ----------- 0 142 0 1 156 156 2 69 138 3 27 81 4 5 20 5 1 5 Total N = 400 Σfx = 400 $$\lambda = \frac{400}{400} = 1$$ $$...
Full solved answer →The rate of denying to take vaccine for COVID19 in a rural population of India is reported to be 0.45 per 10,000 people. If the distribution of denying follows Poisson distribution, what is the probability that in the next 10,000 people, there will be: a.)No one will deny to take vaccine? b.)At least two persons will deny to take vaccine? [5]
- Rate of vaccine denial = 0.45 per 10,000 people - Distribution: Poisson - Sample size for prediction = 10,000 people - Therefore, $\lambda = 0.45$ $$P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}$$ $$P(X=0) = \frac{e^{-0.45}(0.45)^0}{0!} = e^{-0.45}$$ Comput...
Full solved answer →Normal distribution properties and applications
Under which situation Normal distribution will be used. The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $20,000. (a) What is the probability of people earns less than $40,000? (b) What is the probability of people earns between $45,000 and $65,000? (c) What is the probability of people earns more than $70,000?[10]
- Distribution: Normal (approximately) - Mean $\mu = 50{,}000$ - Standard deviation $\sigma = 20{,}000$ - Required: - (a) $P(X < 40{,}000)$ - (b) $P(45{,}000 < X < 65{,}000)$ - (c) $P(X 70{,}000)$ --- The normal distribution is applied when: 1. The variable...
Full solved answer →Define normal distribution and standard normal distribution. In an examination, 10% students got less than 20 marks and 95% got less than 75 marks. Assuming the distribution to be normal, find the mean and standard deviation.[10]
- $P(X < 20) = 0.10$ (10% got less than 20 marks) - $P(X < 75) = 0.95$ (95% got less than 75 marks) - Distribution is normal with unknown mean $\mu$ and standard deviation $\sigma$ --- A normal distribution is a continuous probability distribution that is s...
Full solved answer →Discuss the measure properties of normal distribution. The burning time of an experimental rocket is a random variable having the normal distribution with mean 4.76 seconds and standard deviation 0.04 second respectively. What is probability that this kind of rocket will burn (i) Less than 4.68 seconds (ii) More than 4.80 seconds (iii) Anywhere from 4.70 to 4.82 seconds?μ=4.76,σ=0.04\mu = 4.76, \quad \sigma = 0.04μ=4.76,σ=0.04P(X<4.68), P(X>4.80), P(4.70<X<4.82)P(X < 4.68), ; P(X > 4.80), ; P(4.70 < X < 4.82)P(X<4.68),P(X>4.80),P(4.70<X<4.82)[10]
1. Symmetry: The curve is perfectly symmetric about the mean $\mu$; the two halves are mirror images. 2. Bell-shaped: The probability density function is bell-shaped with its peak at $x = \mu$. 3. Mean = Median = Mode = $\mu$: All three measures of central ...
Full solved answer →A set of final examination grades in Basic Statistics, is following normal distribution with mean of 73 and standard deviation of 8. a.) What is the probability of a student secured less than 93 marks? b.)What is the probability of a student secured marks between 65 and 89? [5]
- Mean $\mu = 73$ - Standard Deviation $\sigma = 8$ - Distribution: Normal Z-score formula: $Z = \dfrac{X - \mu}{\sigma}$ --- Calculate Z for $X = 93$: $$Z = \frac{93 - 73}{8} = \frac{20}{8} = 2.5$$ From the standard normal table: $$P(X < 93) = P(Z < 2.5) =...
Full solved answer →Continuous probability distributions
Suppose a continuous random variable X has the density function. Find; (i) value of k, (ii) P (0 < x < 0.5), and, (iii) E(X) and (iv) E(2X + 4)
$$f(x) = \begin{cases} k(1-x)^2; & \text{for } 0 < x < 1 \ 0; & \text{elsewhere} \end{cases}$$
[5]
- Density: $f(x) = k(1-x)^2$ for $0 < x < 1$; $f(x) = 0$ elsewhere - Required: (i) $k$, (ii) $P(0<X<0.5)$, (iii) $E(X)$, (iv) $E(2X+4)$ --- Total probability must equal 1: $$\int{0}^{1} k(1-x)^2\,dx = 1$$ $$k\left[-\frac{(1-x)^3}{3}\right]{0}^{1} = k\left[0...
Full solved answer →Make Unit 6 stick
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