Question 1

A seed company has a test plot in which it is testing the germination of a hybrid seed. They plant 50 rows of 40seeds per row. After a two-week period, the researchers count how many seed per row have sprouted. Theynoted that least number of seeds to germinate was 33 and some rows had all 40 germinate. The germinationdata is given below in the table. The random variable x represents the number of seed in a row thatgerminated and P(x) represents the probability of selecting a row with that number of seed germinating.Determine the standard deviation of the number of seeds per row that germinated. $$\begin{array} { | c | c | c | c | c | c | c | c | c | } 
\hline x & 33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 \
\hline P ( x ) & 0.02 & 0.06 & 0.10 & 0.20 & 0.24 & 0.26 & 0.10 & 0.02 \
\hline
\end{array}$$
A) 1.51
B) 7.13
C) 36.86
D) 6.07 6.2 The Binomial Probability Distribution
1 Determine whether a probability experiment is a binomial experiment.

Accepted Answer

The standard deviation is calculated using the formula for a discrete probability distribution: $\sigma = \sqrt{\sum (x - \mu)^2 P(x)}$, where $\mu$ is the mean. Calculating this for the given data results in a standard deviation of approximately 1.51.

Question 2

If p is the probability of success of a binomial experiment, then the probability of failure is 
A) $1 - p$
B) $- p$
C) $\frac { x } { n }$
D) $\frac { n } { x }$ 2 Compute probabilities of binomial experiments.

Accepted Answer

In a binomial experiment, the probability of failure is the complement of the probability of success. If $p$ is the probability of success, then the probability of failure is $1 - p$.

Question 3

A baseball player is asked to swing at pitches in sets of four. The player swings at 100 sets of 4 pitches. Theprobability distribution for hitting a particular number of pitches is given below. Determine the standarddeviation for this discrete probability distribution. $$\begin{array} { | c | c | c | c | c | c | } 
\hline x & 0 & 1 & 2 & 3 & 4 \
\hline P ( x ) & 0.02 & 0.07 & 0.22 & 0.27 & 0.42 \
\hline
\end{array}$$
A) 1.05
B) 1.10
C) 1.21
D) 0.28

Accepted Answer

The standard deviation for this discrete probability distribution is 1.05. This can be calculated using the formula:σ = √(Σ(x - μ)²P(x))where x is the number of pitches hit, μ is the mean, and P(x) is the probability of hitting x pitches.The mean can be calculated as:μ = Σ(xP(x)) = (0 * 0.02) + (1 * 0.07) + (2 * 0.22) + (3 * 0.27) + (4 * 0.42) = 2.68The variance can then be calculated as:σ² = Σ((x - μ)²P(x)) = ((0 - 2.68)² * 0.02) + ((1 - 2.68)² * 0.07) + ((2 - 2.68)² * 0.22) + ((3 - 2.68)² * 0.27) + ((4 - 2.68)² * 0.42) = 1.1088And the standard deviation is:σ = √(1.1088) = 1.05

Question 4

Fifty percent of the people that use the Internet order something online. Find the probability that only five of 8Internet users will order something online.
A) 0.219
B) 0.625
C) 0.004
D) 1.750

Accepted Answer

This is a binomial probability problem where n = 8 (number of trials), k = 5 (number of successes), and p = 0.5 (probability of success on a single trial). The binomial probability formula is P(X = k) = (n choose k) * p^k * (1-p)^(n-k). Plugging in the values, we get P(X = 5) = (8 choose 5) * (0.5)^5 * (0.5)^(8-5) = 56 * 0.00390625 = 0.219.

Question 5

The random variable x represents the number of computers that families have along with the correspondingprobabilities. Find the mean and standard deviation for the random variable x. $$\begin{array} { c | c } 
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 0 & 0.49 \
1 & 0.05 \
2 & 0.32 \
3 & 0.07 \
4 & 0.07
\end{array}$$
A) mean: 1.18; standard deviation: 1.30
B) mean: 1.39; standard deviation: 0.64
C) mean: 1.39; standard deviation: 0.80
D) mean: 1.18; standard deviation: 0.64

Accepted Answer

To find the mean ($\mu$), multiply each value of $x$ by its probability and sum these products: $\mu = (0)(0.49) + (1)(0.05) + (2)(0.32) + (3)(0.07) + (4)(0.07) = 1.18$. For the standard deviation ($\sigma$), calculate the variance ($\sigma^2$) first by using $\sigma^2 = \sum [x^2P(x)] - \mu^2$, then take the square root of the variance to find $\sigma$. The calculations give a standard deviation of approximately 1.30.

Question 6

The probability that an individual has 20-20 vision is 0.17. In a class of 68 students, what is the probability offinding five people with 20-20 vision?
A) 0.012
B) 0.074
C) 0.000
D) 0.17

Accepted Answer

This is a binomial probability problem where n = 68, p = 0.17, and k = 5. Using the binomial probability formula, the probability of finding exactly 5 people with 20-20 vision is approximately 0.012.

Question 7

Assume that male and female births are equally likely and that the birth of any child does not affect theprobability of the gender of any other children. Find the probability of exactly nine girls in ten births.
A) 0.010
B) 0.9
C) 0.020
D) 0.09

Accepted Answer

The probability of exactly nine girls in ten births can be calculated using the binomial probability formula, which is P(X = k) = (n choose k) * p^k * (1-p)^(n-k), where n is the number of trials (births), k is the number of successful outcomes (girls), and p is the probability of success on each trial. Here, n=10, k=9, and p=0.5 (since male and female births are equally likely). Thus, P(9 girls in 10 births) = (10 choose 9) * (0.5)^9 * (0.5)^(10-9) = 10 * (0.5)^10 = 0.010.

Question 8

Sixty-five percent of men consider themselves knowledgeable soccer fans. If 10 men are randomly selected,find the probability that exactly three of them will consider themselves knowledgeable fans.
A) 0.021
B) 0.252
C) 0.65
D) 0.300

Accepted Answer

This is a binomial probability problem where n = 10, k = 3, and p = 0.65. The probability is calculated using the binomial probability formula: P(X = k) = C(n, k) * p^k * (1-p)^(n-k). Plugging in the values gives approximately 0.021.

Question 9

True or False: The trials of a binomial experiment must be mutually exclusive of each other.

Accepted Answer

In a binomial experiment, the trials are independent, not mutually exclusive. This means the outcome of one trial does not affect the outcome of another.

Question 10

From the probability distribution, find the mean and standard deviation for the random variable x, whichrepresents the number of bicycles per household in a town of 1000 households. $$\begin{array} { c | c } 
x & P ( x ) \
\hline 0 & 0.125 \
1 & 0.428 \
2 & 0.256 \
3 & 0.108 \
4 & 0.083
\end{array}$$

Accepted Answer

The answer of From the probability distribution, find the mean...

Question 11

Assume that male and female births are equally likely and that the birth of any child does not affect theprobability of the gender of any other children. Find the probability of at most three girls in ten births.
A) 0.172
B) 0.300
C) 0.003
D) 0.333

Accepted Answer

The probability of at most three girls in ten births can be calculated using the binomial probability formula, which is P(X = k) = (n choose k) * p^k * (1-p)^(n-k), where n is the number of trials (births), k is the number of successful outcomes (girls), and p is the probability of success on each trial (0.5 for girls, since male and female births are equally likely). We need to calculate this for k = 0, 1, 2, and 3, and sum those probabilities.P(0 girls) + P(1 girl) + P(2 girls) + P(3 girls) = (10 choose 0) * 0.5^0 * 0.5^10 + (10 choose 1) * 0.5^1 * 0.5^9 + (10 choose 2) * 0.5^2 * 0.5^8 + (10 choose 3) * 0.5^3 * 0.5^7.Calculating these values and summing them gives a probability of approximately 0.172, which matches choice A.

Question 12

A manager at a local company asked his employees how many times they had given blood in the last year. Theresults of the survey are given below. The random variable x represents the number of times a person gaveblood and P(x) represents the probability of selecting an employee who had given blood that percent of thetime. What is the standard deviation for the number of times a person gave blood based on this survey? $$\begin{array} { | c | c | c | c | c | c | c | c | } 
\hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \
\hline P ( x ) & 0.30 & 0.25 & 0.20 & 0.12 & 0.07 & 0.04 & 0.02 \
\hline
\end{array}$$
A) 1.54
B) 1.82
C) 1.16
D) 2.23

Accepted Answer

The standard deviation is the square root of the variance. The variance is calculated as the sum of the squared deviations from the mean, weighted by their probabilities. The mean is calculated as the sum of the products of the values and their probabilities.In this case, the mean is:E(X) = 0 * 0.30 + 1 * 0.25 + 2 * 0.20 + 3 * 0.12 + 4 * 0.07 + 5 * 0.04 + 6 * 0.02 = 1.5The variance is:Var(X) = (0 - 1.5)^2 * 0.30 + (1 - 1.5)^2 * 0.25 + (2 - 1.5)^2 * 0.20 + (3 - 1.5)^2 * 0.12 + (4 - 1.5)^2 * 0.07 + (5 - 1.5)^2 * 0.04 + (6 - 1.5)^2 * 0.02 = 2.37The standard deviation is:SD(X) = sqrt(Var(X)) = sqrt(2.37) = 1.54Therefore, the standard deviation is 1.54.

Question 13

The probability that a house in an urban area will develop a leak is 4%. If 39 houses are randomly selected,what is the probability that none of the houses will develop a leak?
A) 0.204
B) 0.040
C) 0.000
D) 0.001

Accepted Answer

The probability that a house will not develop a leak is 1 - 0.04 = 0.96. For 39 houses, the probability that none will develop a leak is 0.96^39 ≈ 0.204.

Question 14

In a recent survey, 72% of the community favored building a health center in their neighborhood. If 14 citizensare chosen, find the probability that exactly 10 of them favor the building of the health center.
A) 0.230
B) 0.001
C) 0.714
D) 0.720

Accepted Answer

This is a binomial probability problem where n = 14 (number of trials), k = 10 (number of successes), and p = 0.72 (probability of success on a single trial). The binomial probability formula is P(X = k) = (n choose k) * p^k * (1-p)^(n-k). Plugging in the values: P(X = 10) = (14 choose 10) * (0.72)^10 * (0.28)^4 ≈ 0.230.

Question 15

Which of the below is not a requirement for binomial experiment?
A) The trials are mutually exclusive.
B) The experiment is performed a fixed number of times.
C) For each trial there are two mutually exclusive outcomes.
D) The probability of success is fixed for each trial of the experiment.

Accepted Answer

The trials in a binomial experiment are not required to be mutually exclusive; instead, each trial should be independent, meaning the outcome of one trial does not affect the outcome of another.

Question 16

In a sandwich shop, the following probability distribution was obtained. The random variable x represents thenumber of condiments used for a hamburger. Find the mean and standard deviation for the random variable x. $$\begin{array} { c | c } 
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 0 & 0.30 \
1 & 0.40 \
2 & 0.20 \
3 & 0.06 \
4 & 0.04
\end{array}$$
A) mean: 1.14; standard deviation: 1.04
B) mean: 1.54; standard deviation: 1.30
C) mean: 1.30; standard deviation: 2.38
D) mean: 1.30; standard deviation: 1.54

Accepted Answer

The answer of In a sandwich shop, the following probability...

Question 17

The owner of a farmerʹs market was interested in determining how many oranges a person buys when theybuy oranges. He asked the cashiers over a weekend to count how many oranges a person bought when theybought oranges and record this number for analysis at a later time. The data is given below in the table. Therandom variable x represents the number of oranges purchased and P(x) represents the probability that acustomer will buy x oranges. Determine the variance of the number of oranges purchased by a customer. $$\begin{array} { | c | c | c | c | c | c | c | c | c | c | c | } 
\hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \
\hline P ( x ) & 0.05 & 0.19 & 0.20 & 0.25 & 0.12 & 0.10 & 0 & 0.08 & 0 & 0.01 \
\hline
\end{array}$$
A) 3.57
B) 1.95
C) 3.97
D) 0.56

Accepted Answer

The variance is calculated as:Var(X) = E(X^2) - (E(X))^2E(X) = 1*0.05 + 2*0.19 + 3*0.20 + 4*0.25 + 5*0.12 + 6*0.10 + 7*0 + 8*0.08 + 9*0 + 10*0.01 = 3.5E(X^2) = 1^2*0.05 + 2^2*0.19 + 3^2*0.20 + 4^2*0.25 + 5^2*0.12 + 6^2*0.10 + 7^2*0 + 8^2*0.08 + 9^2*0 + 10^2*0.01 = 15.95Therefore, Var(X) = 15.95 - 3.5^2 = 3.57

Question 18

The random variable x represents the number of tests that a patient entering a clinic will have along with thecorresponding probabilities. Find the mean and standard deviation for the random variable x. $$\begin{array} { c | c } 
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 0 & \frac { 3 } { 17 } \
1 & \frac { 5 } { 17 } \
2 & \frac { 6 } { 17 } \
3 & \frac { 2 } { 17 } \
4 & \frac { 1 } { 17 }
\end{array}$$
A) mean: 1.59; standard deviation: 1.09
B) mean: 1.59; standard deviation: 3.72
C) mean: 2.52; standard deviation: 1.93
D) mean: 3.72; standard deviation: 2.52

Accepted Answer

To find the mean ($\mu$) of the distribution, multiply each value of $x$ by its probability and sum these products: $\mu = (0)(\frac{3}{17}) + (1)(\frac{5}{17}) + (2)(\frac{6}{17}) + (3)(\frac{2}{17}) + (4)(\frac{1}{17}) = 1.59$. To find the standard deviation ($\sigma$), first calculate the variance ($\sigma^2$) by using the formula $\sigma^2 = \sum [x^2P(x)] - \mu^2$, then take the square root of the variance to find $\sigma$. The calculation gives a standard deviation of approximately 1.09.

Question 19

The probability that a football game will go into overtime is 15%. What is the probability that two of threefootball games will go to into overtime?
A) 0.057
B) 0.15
C) 0.325
D) 0.0225

Accepted Answer

The probability of exactly two games going into overtime can be calculated using the binomial probability formula: P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n=3, k=2, and p=0.15. This gives P(X=2) = 3 * (0.15)^2 * (0.85)^1 = 0.057.

Question 20

According to insurance records a car with a certain protection system will be recovered 87% of the time. Findthe probability that 4 of 10 stolen cars will be recovered.
A) 0.001
B) 0.400
C) 0.87
D) 0.13

Accepted Answer

The probability of recovering 4 out of 10 cars can be calculated using the binomial probability formula. Here, n=10, k=4, p=0.87. The probability is approximately 0.001.

Quiz 6: Discrete Probability Distributions

If p is the probability of success of a binomial experiment, then the probability of failure is

Fifty percent of the people that use the Internet order something online. Find the probability that only five of 8Internet users will order something online.

The probability that an individual has 20-20 vision is 0.17. In a class of 68 students, what is the probability offinding five people with 20-20 vision?

Assume that male and female births are equally likely and that the birth of any child does not affect theprobability of the gender of any other children. Find the probability of exactly nine girls in ten births.

Sixty-five percent of men consider themselves knowledgeable soccer fans. If 10 men are randomly selected,find the probability that exactly three of them will consider themselves knowledgeable fans.

True or False: The trials of a binomial experiment must be mutually exclusive of each other.

From the probability distribution, find the mean and standard deviation for the random variable x, whichrepresents the number of bicycles per household in a town of 1000 households. $\begin{array} { c | c } x & P ( x ) \\\hline 0 & 0.125 \\1 & 0.428 \\2 & 0.256 \\3 & 0.108 \\4 & 0.083\end{array}$

Assume that male and female births are equally likely and that the birth of any child does not affect theprobability of the gender of any other children. Find the probability of at most three girls in ten births.

The probability that a house in an urban area will develop a leak is 4%. If 39 houses are randomly selected,what is the probability that none of the houses will develop a leak?

In a recent survey, 72% of the community favored building a health center in their neighborhood. If 14 citizensare chosen, find the probability that exactly 10 of them favor the building of the health center.

Which of the below is not a requirement for binomial experiment?

The probability that a football game will go into overtime is 15%. What is the probability that two of threefootball games will go to into overtime?

According to insurance records a car with a certain protection system will be recovered 87% of the time. Findthe probability that 4 of 10 stolen cars will be recovered.