Question 1

Let the random variable x represent the number of heads in five flips of a coin. Construct a table describing the probability distribution, then find the mean and standard deviation.

Accepted Answer

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Question 2

The number of lightning strikes in a year at the top of a particular mountain has a Poisson distribution with a mean of 3.9. Find the probability that in a randomly selected year, the number of lightning strikes is 2.

Accepted Answer

The probability of observing exactly $k$ events in a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean number of events. Here, $\lambda = 3.9$ and $k = 2$. Plugging these values into the formula gives $P(X=2) = \frac{e^{-3.9} \cdot 3.9^2}{2!} = 0.1539$.

Question 3

A mountain search and rescue team receives an average of µ = 0.75 calls per day. Find the probability that on a randomly selected day, they will receive fewer than two calls.

Accepted Answer

The number of calls per day can be modeled by a Poisson distribution with a mean (µ) of 0.75 calls per day. The probability of receiving fewer than two calls (0 or 1 call) is calculated using the Poisson probability formula: P(X < 2) = P(0) + P(1). P(0) = e^(-0.75) * 0.75^0 / 0! and P(1) = e^(-0.75) * 0.75^1 / 1!. Adding P(0) and P(1) gives the probability of receiving fewer than two calls, which is approximately 0.8266.

Question 4

Assume that a probability distribution is described by the discrete random variable x that can assume the values 1, 2, . . . , n; and those values are equally likely. This probability has mean and standard deviation described as follows: $$\mu = \frac { \mathrm { n } + 1 } { 2 } \text { and } \sigma = \sqrt { \frac { \mathrm { n } ^ { 2 } - 1 } { 12 } }$$ Show that the formulas hold for the case of n = 7.

Accepted Answer

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Question 5

The number of calls received by a car towing service averages 21.6 per day (per 24-hour period). After finding the mean number of calls per hour, find the probability that in a randomly selected hour the number of calls is 4.

Accepted Answer

The mean number of calls per hour is 21.6/24 = 0.9. The probability of 4 calls in an hour is given by the Poisson distribution: P(X = 4) = (e^-0.9 * 0.9^4) / 4! = 0.01111.

Question 6

For a certain type of fabric, the average number of defects in each square foot of fabric is 0.3. Find the probability that a randomly selected square foot of the fabric will contain more than one defect.

Accepted Answer

The probability of more than one defect can be found using the Poisson distribution formula. Given the average rate (λ) of 0.3 defects per square foot, the probability of getting more than one defect (k > 1) is 1 minus the probability of getting 0 or 1 defects. Using the Poisson formula, P(k; λ) = (e^(-λ) * λ^k) / k!, for k = 0 and k = 1, we calculate these probabilities and subtract from 1. The calculation for 0 and 1 defects gives us a combined probability, which when subtracted from 1, gives the probability of having more than one defect. The correct calculation aligns with choice A, 0.0369, as the probability of having more than one defect in a square foot of this fabric.

Question 7

Ten apples, four of which are rotten, are in a refrigerator. Three apples are randomly selected without replacement. Let the random variable x represent the number chosen that are rotten. Construct a table describing the probability distribution, then find the mean and standard deviation for the random variable x.

Accepted Answer

The answer of Ten apples, four of which are rotten,...

Question 8

Let the random variable x represent the number of boys in a family of three children. Construct a table describing the probability distribution, then find the mean and standard deviation.

Accepted Answer

The answer of Let the random variable x represent the...

Question 9

If a procedure meets all the conditions of a binomial distribution except that the number of trials is not fixed, then the geometric distribution can be used. The probability of getting the first success on the xth trial is given by Pchoosing a yellow piece of cand x = p 1 - px - 1 , where p is ty in a bag of hard candhe probability of succey is ss on any on0.204. Find the e trial. probAssuame that thbility that e probathe first yellow bility of candy is found in the fourth inspected.

Accepted Answer

The probability of getting the first success on the xth trial is given by P(x) = p(1 - p)^x-1. In this case, p = 0.204 and x = 4, so P(4) = 0.204(1 - 0.204)^3 = 0.103.

Question 10

The Columbia Power Company experiences power failures with a mean of µ = 0.210 per day. Find the probability that there are exactly two power failures in a particular day.

Accepted Answer

The answer of The Columbia Power Company experiences power failures...

Question 11

If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use hypergeometric distribution. If a population has A objects of one type, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n - x objects of type B is $$P ( x ) = \frac { A ! } { ( A - x ) ! x ! } \cdot \frac { B ! } { ( B - n + x ) ! ( n - x ) ! } \div \frac { ( A + B ) ! } { ( A + B - n ) ! n ! }$$ In a relatively easy lottery, a bettor selects 4 numbers from 1 to 14 (without repetition), and a winning 4-number combination is later randomly selected. What is the probability of getting all 4 winning numbers?

Accepted Answer

The answer of If we sample from a small finite...

Question 12

A naturalist leads whale watch trips every morning in March. The number of whales seen has a Poisson distribution with a mean of 4.3. Find the probability that on a randomly selected trip, the number of whales seen is 3.

Accepted Answer

The probability of observing exactly $k$ events in a Poisson distribution is given by $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$, where $\lambda$ is the mean number of events. For $k=3$ and $\lambda=4.3$, $P(X=3) = \frac{4.3^3 e^{-4.3}}{3!} \approx 0.1798$.

Question 13

If the random variable x has a Poisson Distribution with mean µ = 0.604, find the probability that x = 1.

Accepted Answer

The probability that x = 1 for a Poisson distribution is calculated using the formula P(x; µ) = (e^(-µ) * µ^x) / x!, where µ is the mean. For µ = 0.604 and x = 1, P(1; 0.604) = (e^(-0.604) * 0.604^1) / 1! = 0.33016.

Question 14

If the random variable x has a Poisson Distribution with mean µ = 2, find the probability that x = 1.

Accepted Answer

The probability of a Poisson random variable x with mean µ is given by P(x=k) = (e^(-µ) * µ^k) / k!. For x=1 and µ=2, P(x=1) = (e^(-2) * 2^1) / 1! = 0.27067.

Question 15

A computer salesman averages 1.9 sales per week. Use the Poisson distribution to find the probability that in a randomly selected week the number of computers sold is 3.

Accepted Answer

The probability of selling 3 computers in a week is given by the Poisson distribution as:P(X = 3) = (e^-1.9 * 1.9^3) / 3! = 0.1710

Question 16

Write the word or phrase that best completes each statement or answers the question
-An experiment of a gender selection method includes a control group of 8 couples who are not given any treatment intended to influence the genders of their babies. Construct a table listing the possible values of the random variable x (which represents the number of girls among the 8 births)and the corresponding probabilities. Then find the mean and standard deviation for the number of girls in such groups of 10 and the maximum and minimum usual values for the number of girls. Round all results to four decimal places.

Accepted Answer

The answer of Write the word or phrase that best...

Question 17

The binomial distribution applies only to cases involving two types of outcomes, whereas the multinomial distribution involves more than two categories. Suppose we have three types of mutually exclusive outcomes denoted by A, B, and C. Let P A = p1, P B = p2, P C = p3. In n independent trials, the probability of x1 outcomes of type $$A , x _ { 2 }$$ outcomes of type B, and x3 outcomes of type C is given by $$\frac { \mathrm { n } ! } { \left( \mathrm { x } _ { 1 } \right) ! \left( \mathrm { x } _ { 2 } \right) ! \left( \mathrm { x } _ { 3 } \right) ! } \cdot \mathrm { p } _ { 1 } \mathrm { x } _ { 1 } \cdot \mathrm { p } _ { 2 } ^ { \mathrm { x } _ { 2 } } \cdot \mathrm { p } _ { 3 } ^ { \mathrm { x } _ { 3 } }$$ A genetics experiment involves four mutually exclusive genotypes identified as A, B, C, and D, and they are all equally likely. If 13 offspring are tested, find the probability of getting exactly 2 A's,4 B's,2 C's, and 5 D's by expanding the above expression so that it applies to four types of outcomes instead of only three.

Accepted Answer

The answer of The binomial distribution applies only to cases...

Question 18

Assume that x is a random variable in a probability distribution with mean µ and standard deviation ?. Find expressions for the mean and standard deviation if every value of x is modified by first being multiplied by 3, then increased by 5.

Accepted Answer

When a random variable $x$ is transformed into $3x + 5$, the mean $\mu$ is scaled and shifted to $3\mu + 5$, but the standard deviation $\sigma$ is only scaled by the factor of 3, not shifted, because standard deviation is a measure of spread and does not depend on the addition of constants.

Quiz 5: Probability Distributions