Question 1

Obtain the probability distribution of the random variable.
-The following frequency table contains data on home sale prices in the city of Summerhill for the month of June. For a randomly selected sale price between $80,000 and $265,900 let X denote the Number of homes that sold for that price. Find the probability distribution of X. $\begin{array}{c|c}
\text { Sale Price (in thousands) } & \begin{array}{c}
\text { Frequency } \
\text { (No. of homes sold) }
\end{array} \
\hline 80.0-110.9 & 2 \
111.0-141.9 & 5 \
142.0-172.9 & 7 \
173.0-203.9 & 10 \
204.0-234.9 & 3 \
235.0-265.9 & 1
\end{array}$
 
A)
$\begin{array}{c|c}
\text { Sale Price (in thousands) } & \begin{array}{c}
\text { Probability } \
(\mathrm{P}(\mathrm{X}=\mathrm{x})
\end{array} \
\hline 80.0-110.9 & 0.071 \
111.0-141.9 & 0.179 \
142.0-172.9 & 0.250 \
173.0-203.9 & 0.357 \
204.0-234.9 & 0.107 \
235.0-265.9 & 0.360
\end{array}$

B)
$\begin{array}{c|c}
\text { Sale Price (in thousands) } & \begin{array}{c}
\text { Probability } \
(\mathrm{P}(\mathrm{X}=\mathrm{x})
\end{array} \
\hline 80.0-110.9 & 0.071 \
111.0-141.9 & 0.179 \
142.0-172.9 & 0.025 \
173.0-203.9 & 0.357 \
204.0-234.9 & 0.107 \
235.0-265.9 & 0.036
\end{array}$

C)
$\begin{array}{c|c}
\text { Sale Price (in thousands) } & \begin{array}{c}
\text { Probability } \
(\mathrm{P}(\mathrm{X}=\mathrm{x})
\end{array} \
\hline 80.0-110.9 & 0.071 \
111.0-141.9 & 0.197 \
142.0-172.9 & 0.250 \
173.0-203.9 & 0.357 \
204.0-234.9 & 0.107 \
235.0-265.9 & 0.036
\end{array}$

D)
$\begin{array}{c|c}
\text { Sale Price (in thousands) } & \begin{array}{c}
\text { Probability } \
(\mathrm{P}(\mathrm{X}=\mathrm{x})
\end{array} \
\hline 80.0-110.9 & 0.071 \
111.0-141.9 & 0.179 \
142.0-172.9 & 0.250 \
173.0-203.9 & 0.357 \
204.0-234.9 & 0.107 \
235.0-265.9 & 0.036
\end{array}$

Accepted Answer

To find the probability distribution, divide the frequency of homes sold in each price range by the total number of homes sold. The total number of homes sold is $2 + 5 + 7 + 10 + 3 + 1 = 28$. Calculating the probability for each price range: $2/28 = 0.071$, $5/28 = 0.179$, $7/28 = 0.250$, $10/28 = 0.357$, $3/28 = 0.107$, $1/28 = 0.036$. These calculations match the probabilities in option D.

Question 2

Determine the possible values of the random variable.
-The following frequency distribution analyzes the scores on a math test. For a randomly selected score between 40 and 99, let Y denote the number of students with that score on the test. What are The possible values of the random variable Y? 
$\begin{array}{cc}
\hline \text { Scores } & \begin{array}{c}
\text { Number of } \
\text { students }
\end{array} \
\hline 40-59 & 2 \
60-75 & 4 \
76-82 & 6 \
83-94 & 15 \
95-99 & 5
\end{array}$
A)$2,4,6,5$
B) $2,4,6,15,5$
C) $2,4,6,15$
D) 32

Accepted Answer

The possible values of the random variable Y, which denotes the number of students with a specific score range on the test, are directly taken from the frequency distribution: 2, 4, 6, 15, and 5.

Question 3

Find the mean of the Poisson random variable.
-Suppose $X$ has a Poisson distribution with parameter $\lambda = 1.4$. Find the mean of $X$.  
A)$1.4$
B) $1.183$
C) 1
D) $1.96$

Accepted Answer

The mean of a Poisson random variable is equal to its parameter $\lambda$, which in this case is 1.4.

Question 4

Find the expected value of the random variable. Round to the nearest cent unless stated otherwise.
-Sue Anne owns a medium-sized business. Use the probability distribution below, where X describes the number of employees who call in sick on a given day. $$\begin{array} { l | c c c c c } 
\text { Number of Employees Sick } & 0 & 1 & 2 & 3 & 4 \
\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 0.05 & 0.4 & 0.3 & 0.15 & 0.1
\end{array}$$ What is the expected value of the number of employees calling in sick on any given day? Round
The answer to two decimal places. 
A)2.00
B)1.90
C)1.85
D)1.00

Accepted Answer

The answer of Find the expected value of the random...

Question 5

Find the specified probability.
-There are only 8 chairs in our whole house. Whenever there is a party some people have no where to sit. The number of people at our parties (call it the random variable X)changes with each party.
Past records show that the probability distribution of X is as shown in the following table. Find the
Probability that everyone will have a place to sit at our next party. $\begin{array}{cccccccc}
\mathrm{X} & 5 & 6 & 7 & 8 & 9 & 10 & >10 \
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & 0.05 & 0.05 & 0.20 & 0.15 & 0.15 & 0.10 & 0.30
\end{array}$
A)$0.55$
B) $0.45$
C) $0.15$
D) $0.05$

Accepted Answer

The probability that everyone will have a place to sit is the sum of probabilities for X = 5, 6, 7, and 8, which is 0.05 + 0.05 + 0.20 + 0.15 = 0.45.

Question 6

Determine the binomial probability formula given the number of trials and the success probability for Bernoulli trials.Let X denote the total number of successes. Round to three decimal places.
-$$\mathrm { n } = 4 , \mathrm { p } = \frac { 1 } { 4 } , \mathrm { P } ( \mathrm { X } = 3 )$$  
A)$0.070$
B) $0.047$
C) $0.061$
D) $0.016$

Accepted Answer

The answer of Determine the binomial probability formula given the...

Question 7

Use the Poisson Distribution to find the indicated probability. Round to three decimal places when necessary.
-A naturalist leads whale watch trips every morning in March. The number of whales seen has a Poisson distribution with parameter $\lambda$ = 4.1. Find the probability that on a randomly selected trip,
The number of whales seen is 
A)3. 0.19
B)0.324
C)0.381
D)0.571

Accepted Answer

The probability of seeing 3 whales is given by the Poisson distribution with parameter $\lambda$ = 4.1:$$P(X = 3) = \frac{e^{-4.1}4.1^3}{3!} = 0.190$$

Question 8

Find the standard deviation of the Poisson random variable. Round to three decimal places.
-Suppose X has a Poisson distribution with parameter ʎ = 1.300. Find the standard deviation of X. 
A)1.690
B)1.14
C)1.300
D)1.000

Accepted Answer

The standard deviation of a Poisson random variable is the square root of its mean (λ), so for λ = 1.300, the standard deviation is sqrt(1.300) ≈ 1.140.

Question 9

Find the standard deviation of the random variable.
-A police department reports that the probabilities that 0, 1, 2, and 3 burglaries will be reported in a given day are 0.48, 0.36, 0.14, and 0.02, respectively. Find the standard deviation for the probability
Distribution. Round the answer to two decimal places. 
A)0.61
B)1.05
C)1.04
D)0.78

Accepted Answer

The answer of Find the standard deviation of the random...

Question 10

Provide an appropriate response.
-Which of the random variables described below is/are discrete random variables? The random variable X represents the number of heads when a coin is flipped 20 times.The random variable Y represents the number of calls received by a car tow service in a year. The random variable Z represents the weight of a randomly selected student. 
A)X, Y, and Z
B)X and Y
C)X only
D)Y only

Accepted Answer

Discrete random variables take on countable values. X, the number of heads in 20 coin flips, and Y, the number of calls received in a year, are countable. Z, the weight of a student, is continuous since weight can take on any value within a range.

Question 11

Find the specified probability.
-Use the special addition rule and the following probability distribution to determine $P ( X \geq 8 )$.
$\begin{array}{cccccccc}
\mathrm{X} & 5 & 6 & 7 & 8 & 9 & 10 & 11 \
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & 0.05 & 0.05 & 0.20 & 0.15 & 0.15 & 0.10 & 0.30
\end{array}$  
A)$0.45$
B) $0.15$
C) $0.70$
D) $0.30$

Accepted Answer

To find $P(X \geq 8)$, we add the probabilities of X being 8, 9, 10, and 11. Thus, $0.15 + 0.15 + 0.10 + 0.30 = 0.70$.

Question 12

Determine the required probability by using the Poisson approximation to the binomial distribution. Round to threedecimal places.
-The rate of defects among CD players of a certain brand is 1.4%. Use the Poisson approximation to the binomial distribution to find the probability that among 230 such CD players received by a
Store, there is at most one defective. 
A)0.831
B)0.129
C)0.960
D)0.169

Accepted Answer

The answer of Determine the required probability by using the...

Question 13

Provide an appropriate response.
-For any discrete random variable, the possible values of the random variable form a
finite set of numbers.

Accepted Answer

A discrete random variable can have either a finite set or a countably infinite set of possible values.

Question 14

Find the mean of the binomial random variable. Round to two decimal places when necessary.
-The probability that a radish seed will germinate is 0.7. A gardener plants seeds in batches of 12. Find the mean for the random variable X, the number of seeds germinating in each batch. 
A)10.8
B)3.6
C)8.52
D)8.4

Accepted Answer

The answer of Find the mean of the binomial random...

Question 15

Calculate the specified probability
-Suppose that $\mathrm { K }$ is a random variable. Given that $\mathrm { P } ( - 3.45 \leq \mathrm { K } \leq 3.45 ) = 0.35$, and that $\mathrm { P } ( \mathrm { K } < - 3.45 ) =$ $P ( K > 3.45 )$, find $P ( K > 3.45 )$.  
A)$0.35$
B) $0.65$
C) $0.325$
D) $1.725$

Accepted Answer

The total probability is 1. Given $P(-3.45 \leq K \leq 3.45) = 0.35$, the remaining probability is $1 - 0.35 = 0.65$. Since $P(K < -3.45) = P(K > 3.45)$, we divide the remaining probability by 2 to find $P(K > 3.45) = 0.65 / 2 = 0.325$.

Question 16

Find the expected value of the random variable. Round to the nearest cent unless stated otherwise.
-Suppose you buy 1 ticket for $1 out of a lottery of 1,000 tickets where the prize for the one winning ticket is to be $500. What is your expected value? 
A)-$1.00
B)-$0.40
C)-$0.50
D)$0.00

Accepted Answer

The expected value is calculated as the probability of winning times the prize minus the cost of the ticket. The probability of winning is 1/1000, so the expected value is (1/1000) * $500 - $1 = $0.50 - $1 = -$0.50. However, since the options provided do not include negative values, the closest positive expected value is $0.50.

Question 17

Find the mean of the Poisson random variable.
-In one town, the number of burglaries in a week has a Poisson distribution with parameter $\lambda = 1.8$. Let $X$ denote the number of burglaries in the town in a randomly selected week. Find the mean of X.  
A)$1.342$
B) $0.9$
C) $1.8$
D) $3.24$

Accepted Answer

The mean of a Poisson random variable is equal to its parameter $\lambda$, which in this case is 1.8.

Question 18

Find the mean of the binomial random variable. Round to two decimal places when necessary.
-A die is rolled 4 times and the number of times that two shows on the upper face is counted. If this experiment is repeated many times, find the mean for the random variable X, the number of twos. 
A)3.33
B)0.67
C)1.33
D)1

Accepted Answer

The mean of a binomial random variable is given by $ \mu = np $, where $ n $ is the number of trials and $ p $ is the probability of success on a single trial. Here, $ n = 4 $ (since the die is rolled 4 times) and $ p = \frac{1}{6} $ (since the probability of rolling a two on a six-sided die is 1 out of 6). Thus, $ \mu = 4 \times \frac{1}{6} = \frac{4}{6} = 0.67 $.

Question 19

Determine the possible values of the random variable.
-Suppose a coin is tossed four times. Let X denote the total number of tails obtained in the four tosses. What are the possible values of the random variable X? 
A) $1,2,3,4$
B) HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
C) $1,2,3$
D) $0,1,2,3,4$

Accepted Answer

The possible values of the random variable X, which represents the total number of tails obtained in four coin tosses, range from 0 (no tails) to 4 (all tails), inclusive.

Question 20

Use the Poisson Distribution to find the indicated probability. Round to three decimal places when necessary.
-$$\lambda = 0.466 ; \mathrm { P } ( \mathrm { X } = 2 )$$  
A)$0.006$
B) $0.085$
C) $0.068$
D) $0.173$

Accepted Answer

The Poisson probability of exactly $k$ events in an interval is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$. For $\lambda = 0.466$ and $k = 2$, $P(X=2) = \frac{e^{-0.466} \cdot 0.466^2}{2!} \approx 0.068$.

Quiz 5: Discrete Random Variables

Find the mean of the Poisson random variable. -Suppose $X$ has a Poisson distribution with parameter $\lambda = 1.4$ . Find the mean of $X$ .

Determine the binomial probability formula given the number of trials and the success probability for Bernoulli trials.Let X denote the total number of successes. Round to three decimal places. - $\mathrm { n } = 4 , \mathrm { p } = \frac { 1 } { 4 } , \mathrm { P } ( \mathrm { X } = 3 )$

Find the standard deviation of the Poisson random variable. Round to three decimal places. -Suppose X has a Poisson distribution with parameter ʎ = 1.300. Find the standard deviation of X.

Provide an appropriate response. -For any discrete random variable, the possible values of the random variable form a finite set of numbers.

Find the mean of the binomial random variable. Round to two decimal places when necessary. -The probability that a radish seed will germinate is 0.7. A gardener plants seeds in batches of 12. Find the mean for the random variable X, the number of seeds germinating in each batch.

Calculate the specified probability -Suppose that $\mathrm { K }$ is a random variable. Given that $\mathrm { P } ( - 3.45 \leq \mathrm { K } \leq 3.45 ) = 0.35$ , and that $\mathrm { P } ( \mathrm { K } < - 3.45 ) =$ $P ( K > 3.45 )$ , find $P ( K > 3.45 )$ .

Find the expected value of the random variable. Round to the nearest cent unless stated otherwise. -Suppose you buy 1 ticket for $1 out of a lottery of 1,000 tickets where the prize for the one winning ticket is to be $500. What is your expected value?

Find the mean of the Poisson random variable. -In one town, the number of burglaries in a week has a Poisson distribution with parameter $\lambda = 1.8$ . Let $X$ denote the number of burglaries in the town in a randomly selected week. Find the mean of X.

Find the mean of the binomial random variable. Round to two decimal places when necessary. -A die is rolled 4 times and the number of times that two shows on the upper face is counted. If this experiment is repeated many times, find the mean for the random variable X, the number of twos.

Determine the possible values of the random variable. -Suppose a coin is tossed four times. Let X denote the total number of tails obtained in the four tosses. What are the possible values of the random variable X?

Use the Poisson Distribution to find the indicated probability. Round to three decimal places when necessary. - $\lambda = 0.466 ; \mathrm { P } ( \mathrm { X } = 2 )$