Question 1

Use the given values of  n=93  and  p=0.24  to find the minimum value that is not significantly low, $\mu - 2 \sigma$ , and the maximum value that is not significantly high, $\mu + 2 \sigma$ Round your answer to the nearest hundredth unless otherwise noted.
A) Minimum:  30.56 ;  maximum:  14.08 
B) Minimum: 14.08; maximum:  30.56 
C) Minimum: 18.2; maximum:  26.44 
D) Minimum:  -11.61 ;  maximum:  56.25

Accepted Answer

For a binomial distribution, $\mu = np$ and $\sigma = \sqrt{np(1-p)}$. Here, $\mu = 93 \times 0.24 = 22.32$ and $\sigma = \sqrt{93 \times 0.24 \times 0.76} \approx 4.12$. Thus, $\mu - 2\sigma \approx 22.32 - 2 \times 4.12 = 14.08$ and $\mu + 2\sigma \approx 22.32 + 2 \times 4.12 = 30.56$.

Question 2

The braking time of a car. Identify the given random variable as being discrete or continuous
A) Discrete
B) Continuous

Accepted Answer

The braking time of a car is a continuous random variable because it can take on any value within an interval, including fractions of seconds.

Question 3

Determine whether the given procedure results in a binomial distribution. If not, state the reason why. Spinning a roulette wheel 7 times, keeping track of the winning numbers.
A) Not binomial: the trials are not independent.
B) Not binomial: there are more than two outcomes for each trial.
C) Not binomial: there are too many trials.
D) The procedure results in a binomial distribution.

Accepted Answer

The procedure does not result in a binomial distribution because each spin of a roulette wheel can result in one of many outcomes (more than two), not just success or failure, which is a requirement for a binomial distribution.

Question 4

Mars, Inc. claims that  20%  of its M&M plain candies are orange. A sample of 100 such candies is randomly selected. Find the mean and standard deviation for the number of orange candies in such groups of 100 .
A) $\mu = 20 , \sigma = 4.0$
B) $\mu = 20 , \sigma = .20$
C) $\mu = .020 , \sigma = .20$
D) $\mu = .20 , \sigma = 4.0$

Accepted Answer

The mean ($\mu$) of orange candies in a sample of 100 is found using the formula $\mu = np$, where $n = 100$ (the sample size) and $p = 0.20$ (the probability of getting an orange candy). Thus, $\mu = 100 \times 0.20 = 20$. The standard deviation ($\sigma$) is found using the formula $\sigma = \sqrt{np(1-p)}$, which gives $\sigma = \sqrt{100 \times 0.20 \times (1-0.20)} = \sqrt{16} = 4.0$.

Question 5

A die is rolled nine 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 number of twos.
A) 7.5 twos
B) 1.5 twos
C) 2.25 twos
D) 3 twos

Accepted Answer

The mean number of twos in nine rolls of a die can be calculated using the formula for the expected value of a binomial distribution, which is n*p, where n is the number of trials (9 rolls) and p is the probability of getting a two in a single roll (1/6). Therefore, the mean is 9*(1/6) = 1.5 twos.

Question 6

The accompanying table shows the probability distribution for x, the number that shows up when a loaded die is rolled. $\begin{array}{c|c}
\boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) \
\hline 1 & 0.14 \
2 & 0.16 \
3 & 0.12 \
4 & 0.14 \
5 & 0.13 \
6 & 0.31
\end{array}$

A) $\mu = 3.9$
B) $\mu = 0.2$
C) $\mu = 3.5$
D) $\mu = 3.8$

Accepted Answer

The mean ($\mu$) of a probability distribution is calculated by summing the products of each value of $x$ and its corresponding probability $P(x)$. For this distribution, $\mu = (1)(0.14) + (2)(0.16) + (3)(0.12) + (4)(0.14) + (5)(0.13) + (6)(0.31) = 3.9$.

Question 7

Match the following values as a A)discrete random variable, B)continuous random variable, or C)not a random variable:
Exact weight of quarters now in circulation in the United States ______________
Shoe sizes of humans ____________________
Political party affiliations of adults in the United States _________________
A)B, A, C
B)C, A, B
C)A, B, C
D)A, C, B

Accepted Answer

The exact weight of quarters is a continuous random variable as it can take any value within a range. Shoe sizes are discrete because they are countable and distinct. Political party affiliations are not numerical and thus not a random variable.

Question 8

In a certain town, 22% of voters favor a given ballot measure. For groups of 21 voters, find the variance for the number who favor the measure. 
A) $3.6$ voters $^ { 2 }$
B) $4.6$ voters $^ { 2 }$
C) $1.9$ voters $^ { 2 }$
D) 13 voters $^ { 2 }$

Accepted Answer

The variance for a binomial distribution is given by the formula $\sigma^2 = np(1-p)$, where $n$ is the number of trials (in this case, voters, which is 21) and $p$ is the probability of success (in this case, the probability that a voter favors the measure, which is 0.22). Plugging in the values, we get $\sigma^2 = 21(0.22)(1-0.22) = 21(0.22)(0.78) = 3.6$ voters$^2$.

Question 9

Find the mean, $\mu$ for the binomial distribution which has the stated values of  n and p. Round answer to the nearest tenth.  n=676 ; p=0.7 
A) $\mu = 474.9$
B) $\mu = 473.2$
C) $\mu = 474.5$
D) $\mu = 471.7$

Accepted Answer

The answer of Find the mean, $\mu$ for the binomial...

Question 10

Find the standard deviation,  $\sigma$, for the binomial distribution which has the stated values of n and p . Round your answer to the nearest hundredth.
 n=503 ; p=0.7 
A) $\sigma = 14.40$
B) $\sigma = 10.28$
C) $\sigma = 7.87$
D) $\sigma = 13.55$

Accepted Answer

The answer of Find the standard deviation,  $\sigma$, for...

Question 11

Use the given values of  n=2112 , $p = \frac { 3 } { 4 }$ to find the minimum value that is not significantly low, $\mu - 2 \sigma$ and the maximum value that is not significantly high, $\mu + 2 \sigma$ Round your answer to the nearest hundredth unless otherwise noted.
A) Minimum:  1478.7 ;  maximum:  1689.3 
B) Minimum:  1544.2 ;  maximum:  1623.8 
C) Minimum:  1564.1 ;  maximum:  1603.9 
D) Minimum: 1623.8; maximum:  1544.2

Accepted Answer

The mean, $\mu$, of a binomial distribution is given by $\mu = np$, and the standard deviation, $\sigma$, is given by $\sigma = \sqrt{np(1-p)}$. For $n = 2112$ and $p = \frac{3}{4}$, $\mu = 2112 \times \frac{3}{4} = 1584$, and $\sigma = \sqrt{2112 \times \frac{3}{4} \times \frac{1}{4}} = \sqrt{396}$ which is approximately 19.9. Therefore, $\mu - 2\sigma \approx 1584 - 2 \times 19.9 = 1544.2$ and $\mu + 2\sigma \approx 1584 + 2 \times 19.9 = 1623.8$.

Question 12

On a multiple choice test with 17 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the mean for the number of correct answers.
A) 8.5 questions
B) 12.8 questions
C) 4.3 questions
D) 5.7 questions

Accepted Answer

Since there are four possible answers for each question and only one of them is correct, the probability of guessing the correct answer for any question is 1/4. For 17 questions, the mean (expected value) of correct answers is 17 * (1/4) = 4.25, which rounds to 4.3 questions.

Question 13

Find the standard deviation, $\sigma$, for the binomial distribution which has the stated values of  n  and  p . Round your answer to the nearest hundredth.
 n=38 ; p=2 / 5 
A) $\sigma = 0.61$
B) $\sigma = 6.29$
C) $\sigma = 3.02$
D) $\sigma = 7.14$

Accepted Answer

The standard deviation for a binomial distribution is calculated using the formula $\sigma = \sqrt{n \cdot p \cdot (1-p)}$. Substituting the given values, $\sigma = \sqrt{38 \cdot \frac{2}{5} \cdot \frac{3}{5}} = \sqrt{9.12} \approx 3.02$.

Question 14

Find the standard deviation, $\sigma$ , for the binomial distribution which has  n=48  and $p = 3 / 5$ Round your answer to the nearest hundredth.
A) $\sigma = 6.66$
B) $\sigma = 0.98$
C) $\sigma = 7.51$
D) $\sigma = 3.39$

Accepted Answer

The standard deviation for a binomial distribution is calculated using the formula $\sigma = \sqrt{n \cdot p \cdot (1-p)}$. Substituting the given values, $\sigma = \sqrt{48 \cdot \frac{3}{5} \cdot \frac{2}{5}} = \sqrt{48 \cdot 0.6 \cdot 0.4} = \sqrt{11.52} \approx 3.39$.

Question 15

Suppose that weight of adolescents is being studied by a health organization and that the accompanying tables describes the probability distribution for three randomly selected adolescents, where x is the number who are considered morbidly obese. Is it significant to
Have no obese subjects among three randomly selected adolescents? $\begin{array}{c|c}
\boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) \
\hline 0 & 0.111 \
1 & 0.215 \
2 & 0.450 \
3 & 0.224
\end{array}$

A) $\mathrm { Yes }$
B) No

Accepted Answer

The probability of having no obese subjects (x=0) is 0.111, which is not extremely low. Typically, a probability below 0.05 is considered significant, so 0.111 is not significant.

Question 16

A tennis player makes a successful first serve  51%  of the time. If she serves 9 times, what is the probability that she gets exactly 3 successful first serves in? Assume that each serve is independent of the others.
A)  0.154 
B)  0.00184 
C)  0.133 
D)  0.0635

Accepted Answer

This is a binomial probability problem where the number of trials n = 9, the number of successes k = 3, and the probability of success on a single trial p = 0.51. The probability of exactly k successes in n trials is given by the formula P(X = k) = (n choose k) * p^k * (1-p)^(n-k). Plugging in the values, we get P(X = 3) ≈ 0.154.

Question 17

The Acme Candy Company claims that  60%  of the jawbreakers it produces weigh more than  0.4  ounces. Suppose that 800 jawbreakers are selected at random from the production lines. Would it be significant for this sample of 800 to contain 494 jawbreakers that weigh more than  0.4  ounces? Consider as significant any result that differs from the mean by more than 2 standard deviations. That is, significant values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$
A) Yes
B) No

Accepted Answer

Given that 60% of the jawbreakers are expected to weigh more than 0.4 ounces, the mean ($\mu$) of jawbreakers weighing more than 0.4 ounces in a sample of 800 would be $800 \times 0.6 = 480$. The standard deviation ($\sigma$) for a binomial distribution is calculated using the formula $\sigma = \sqrt{n \times p \times (1-p)}$, where $n$ is the sample size and $p$ is the probability of success. Substituting the given values, we get $\sigma = \sqrt{800 \times 0.6 \times 0.4} = \sqrt{192} \approx 13.86$. The range for significant values more than 2 standard deviations from the mean would be $\mu \pm 2\sigma$, which calculates to $480 \pm 2 \times 13.86$, or $452.28$ to $507.72$. Since 494 jawbreakers fall within this range, the result is not considered significant.

Question 18

Assume that a procedure yields a binomial distribution with a trial repeated  n=30  times. Use the binomial probability formula to find the probability of  x=5  successes given the probability  p=1/5  of success on a single trial. Round to three decimal places.
A)  0.172 
B)  0.067 
C)  0.198 
D)  0.421

Accepted Answer

The binomial probability formula is $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $\binom{n}{k}$ is the binomial coefficient. Plugging in $n=30$, $k=5$, and $p=1/5$, we get $P(X = 5) = \binom{30}{5}(1/5)^5(4/5)^{25}$. Calculating this gives approximately 0.172.

Question 19

The probability that a call received by a certain switchboard will be a wrong number is 0.02. Use the Poisson distribution to approximate the probability that among 150 calls received by the switchboard, there are at least two wrong numbers. Round your answer to four
Decimal places.
A) 0.8009
B) 0.7760
C) 0.2240
D) 0.1991
E) 0.5768

Accepted Answer

The Poisson distribution can be used to approximate the probability of a given number of events happening in a fixed interval of time or space if these events happen with a known constant rate and independently of the time since the last event. The probability of exactly $k$ events in an interval is given by the formula $P(k) = \frac{\lambda^k e^{-\lambda}}{k!}$, where $\lambda$ is the average rate (mean) of events per interval, and $e$ is the base of the natural logarithm (approximately equal to 2.71828).Given that the probability of a call being a wrong number is 0.02, and there are 150 calls, the expected number of wrong numbers ($\lambda$) is $150 \times 0.02 = 3$.To find the probability of at least two wrong numbers, it's easier to calculate the complement, which is the probability of 0 or 1 wrong number, and subtract it from 1.$$P(0) = \frac{3^0 e^{-3}}{0!} = e^{-3}$$$$P(1) = \frac{3^1 e^{-3}}{1!} = 3e^{-3}$$The probability of 0 or 1 wrong numbers is $P(0) + P(1) = e^{-3} + 3e^{-3} = 4e^{-3}$.The probability of at least two wrong numbers is $1 - (P(0) + P(1)) = 1 - 4e^{-3}$.Calculating this gives a value close to 0.8009, making choice A the correct answer.

Question 20

In a Gallup poll of randomly selected adults, 66% said that they worry about identity theft. For a group of 1013 adults, the mean of those who do not worry about identify theft is closest to:
A) 34 adults
B) 66 adults
C) 669 adults
D) 344 adults

Accepted Answer

If 66% worry about identity theft, then 34% do not worry about it. For 1013 adults, 34% of 1013 is approximately 344 (0.34 * 1013 ≈ 344.42).

Quiz 5: Discrete Probability Distributions

Use the given values of n=93 and p=0.24 to find the minimum value that is not significantly low, $\mu - 2 \sigma$ , and the maximum value that is not significantly high, $\mu + 2 \sigma$ Round your answer to the nearest hundredth unless otherwise noted.

The braking time of a car. Identify the given random variable as being discrete or continuous

Determine whether the given procedure results in a binomial distribution. If not, state the reason why. Spinning a roulette wheel 7 times, keeping track of the winning numbers.

Mars, Inc. claims that 20% of its M&M plain candies are orange. A sample of 100 such candies is randomly selected. Find the mean and standard deviation for the number of orange candies in such groups of 100 .

A die is rolled nine 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 number of twos.

The accompanying table shows the probability distribution for x, the number that shows up when a loaded die is rolled. $\begin{array}{c|c}\boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) \\\hline 1 & 0.14 \\2 & 0.16 \\3 & 0.12 \\4 & 0.14 \\5 & 0.13 \\6 & 0.31\end{array}$

Match the following values as a A) discrete random variable, B) continuous random variable, or C) not a random variable: Exact weight of quarters now in circulation in the United States Shoe sizes of humans Political party affiliations of adults in the United States ___

In a certain town, 22% of voters favor a given ballot measure. For groups of 21 voters, find the variance for the number who favor the measure.

Find the mean, $\mu$ for the binomial distribution which has the stated values of n and p. Round answer to the nearest tenth. n=676 ; p=0.7

Find the standard deviation, $\sigma$ , for the binomial distribution which has the stated values of n and p . Round your answer to the nearest hundredth. n=503 ; p=0.7

Use the given values of n=2112 , $p = \frac { 3 } { 4 }$ to find the minimum value that is not significantly low, $\mu - 2 \sigma$ and the maximum value that is not significantly high, $\mu + 2 \sigma$ Round your answer to the nearest hundredth unless otherwise noted.

On a multiple choice test with 17 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the mean for the number of correct answers.

Find the standard deviation, $\sigma$ , for the binomial distribution which has the stated values of n and p . Round your answer to the nearest hundredth. n=38 ; p=2 / 5

Find the standard deviation, $\sigma$ , for the binomial distribution which has n=48 and $p = 3 / 5$ Round your answer to the nearest hundredth.

A tennis player makes a successful first serve 51% of the time. If she serves 9 times, what is the probability that she gets exactly 3 successful first serves in? Assume that each serve is independent of the others.

Assume that a procedure yields a binomial distribution with a trial repeated n=30 times. Use the binomial probability formula to find the probability of x=5 successes given the probability p=1/5 of success on a single trial. Round to three decimal places.

The probability that a call received by a certain switchboard will be a wrong number is 0.02. Use the Poisson distribution to approximate the probability that among 150 calls received by the switchboard, there are at least two wrong numbers. Round your answer to four Decimal places.

In a Gallup poll of randomly selected adults, 66% said that they worry about identity theft. For a group of 1013 adults, the mean of those who do not worry about identify theft is closest to: