CH 2-6 Statistics

According to a poll, 37 out of every 100 people support a new bill about to be signed into law. What is the probability of selecting a person that is in favor of this new bill?

0.37

Ronald finished his daily jog and wants a drink. In his refrigerator, there are 2 cans of energy drink, 9 cans of soda, and 11 bottles of water. If Ronald chooses a drink at random, what is the probability that he chooses a bottle of water? Give your answer as a fraction.

1/2

The amount of time it takes John to wait in line at the bank is continuous and uniformly distributed between 7 minutes and 17 minutes. What is the probability that it takes John between 9 and 12 minutes to wait in line at the bank? 0.050 0.300 0.400 0.600 0.700

0.300

Key Terms

Bernoulli trial: an experiment with a set of identical trials each with only two possible outcomes. Bernoulli trial is also known as a binomial trial Binomial distribution: a common discrete probability function that includes a a fixed number of identical trials in an experiment where the only possible outcomes are success and failure. Binomial distribution is sometimes known as Bernoulli distribution

Karl and Fredo are basketball players who want to find out how they compare to their team in points per game. The mean amount of points per game and standard deviations for their team were calculated. Karl's z-score is 0.9. Fredo's z-score is −0.65. Which of the following statements are true about how Karl and Fredo compare to their team in points per game? Select all that apply. The given distribution is not a probability distribution, since the sum of probabilities is not equal to 1. The given distribution is not a probability distribution, since at least one of the probabilities is greater than 1 or less than 0. The given distribution is a probability distribution, since the probabilities lie inclusively between 0 and 1. The given distribution is a probability distribution, since the sum of probabilities is equal to 1

Karl's average points per game is 0.9 standard deviations greater than his teammates' average points per game. Fredo's average points per game is closer to the team's mean than Karl's.

A hospital takes record of any birth that occurs there every day. On one day, the hospital reports that 35 of the 62 babies born were girls. Assuming that all of the parents did not have any gender selection procedures, there is a probability of 0.31 of getting these results by chance. Do these results have statistical significance at the 0.05 level of significance? = Yes, the probability of the results occurring is less than 0.05. No, the probability of the results occurring is less than 0.05. Yes, the probability of the results occurring is greater than 0.05. No, the probability of the results occurring is greater than 0.05.

No, the probability of the results occurring is greater than 0.05.

Suppose there is an estate sale featuring many items sold at different prices. The mean selling price of the items is 75 dollars, and the standard deviation is 20. A piece of costume jewelry is priced at 15 dollars. Which of the following statements is true? The price of the costume jewelry is 6 standard deviations to the right of the mean. The price of the costume jewelry is 3 standard deviations to the right of the mean. The price of the costume jewelry is 6 standard deviations to the left of the mean. The price of the costume jewelry is 3 standard deviations to the left of the mean.

The price of the costume jewelry is 3 standard deviations to the left of the mean.

A standard roulette wheel with slots labeled 0, 00, 1, 2, 3, ... , 36 is spun 60 times. Of these spins, the number 7 is spun 7 times. Assuming the wheel is fair, the probability of this occurring by chance is 0.00001. Do these results have statistical significance at the 0.01 level of significance? No, the probability of the results occurring is greater than 0.01. Yes, the probability of the results occurring is less than 0.01. No, the probability of the results occurring is less than 0.01. Yes, the probability of the results occurring is greater than 0.01.

Yes, the probability of the results occurring is less than 0.01. Since the probability of the results occurring, 0.00001, is less than 0.01, the data do have statistical significance at the 0.01 level.

Skewed Data

data that has a main concentration of large values with several other much smaller values, or vice versa - if there is a large concentration of small values with several other much larger values

In a large population, about 45% of people prefer tea over coffee. A researcher takes a random sample of 13 people and surveys whether they prefer tea over coffee. Use the binomial distribution to compute the probability that exactly 6 of the people in the sample prefer tea over coffee. Identify the following information required to find the probability of people who prefer tea over coffee.

n = 13 trials x = 6 successes p(x) = 0.45 probability

Identify the parameters p and n in the following binomial distribution scenario. Jack, a bowler, has a 0.38 probability of throwing a strike and a 0.62 probability of not throwing a strike. If Jack bowls 20 times, he wants to know the probability that he throws more than 10 strikes. (Consider a strike a success in the binomial distribution.) p=0.38,n=0.62 p=0.38,n=10 p=0.38,n=20 p=0.62,n=10 p=0.62,n=20

p=0.38,n=20 In a binomial distribution, there are only two possible outcomes. p denotes the probability of the event or trial resulting in a success. In this scenario, it would be the probability of Jack bowling a strike, which is 0.38. The total number of repeated and identical events or trials is denoted by n. In this scenario, Jack bowls a total of 20 times, so n=20.

The burning times of scented candles, in minutes, are normally distributed with a mean of 249 and a standard deviation of 20. Find the number of minutes a scented candle burns if it burns for a shorter time than 60% of all scented candles. Use a TI-83, TI-83 plus, or TI-84 calculator, and round your answer to two decimal places.

$243.93​ Here, the mean, μ, is 249 and the standard deviation, σ, is 20. In this case, 60% of scented candles burn longer than this scented candle, so the area to the right of x is 60%=0.60. So the area to the left of x is 1−0.60=0.40. The value of x can be calculated using a TI-83, TI-83 plus, or TI-84 calculator. 1. Press 2nd and then VARS for the DISTR menu. 2. Press 3 for invNorm(. 3. Enter 0.40,249,20) and press ENTER. The answer, rounded to two decimal places, is x≈243.93. Thus, a scented candle that burns for a shorter time than 60% of all scented candles burns for 243.93 minutes.

A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour. Find the probability that a household personal computer is used for entertainment between 2.1 and 2.45 hours per day. Round your answer to four decimal places.

0.2367​ Let X represent the random variable. The mean is μ=2, and the standard deviation is σ=0.5. The lower bound is 2.1, and the upper bound is 2.45. Now use a TI-83, TI-83 plus, or TI-84 calculator to find the probability. 1. Press 2nd and VARS for the DISTR menu. 2. Press 2 for normalcdf(. 3. Enter 2.1,2.45,2,0.5) and press ENTER. The answer, rounded to four decimal places, is P(2.1<X<2.45)≈0.2367.

Suppose that the weight of sweet cherries is normally distributed with mean μ=6 ounces and standard deviation σ=1.4 ounces. What proportion of sweet cherries weigh less than 5 ounces? Round your answer to four decimal places.

0.2375 Let X represent the random variable. The mean is μ=6, and the standard deviation is σ=1.4. The lower bound is negative infinity, which is represented by −9999, and the upper bound is 5. Now use a TI-83, TI-83 plus, or TI-84 calculator to find the probability. 1. Press 2nd and VARS for the DISTR menu. 2. Press 2 for normalcdf(. 3. Enter −9999,5,6,1.4) and press ENTER. The answer, rounded to four decimal places, is P(X<5)≈0.2375.

Give the numerical value of the parameter p in the following binomial distribution scenario.A softball pitcher has a 0.675 probability of throwing a strike for each pitch and a 0.325 probability of throwing a ball. If the softball pitcher throws 29 pitches, we want to know the probability that exactly 19 of them are strikes.Consider strikes as successes in the binomial distribution. Do not include p= in your answer.

0.675 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.675.

Suppose that the annual household income in a region is normally distributed with mean $25,000 and standard deviation $6,000. If the poverty level for the region is at or below $12,000, what percentage of the population lives in poverty? Round your answer to two decimal places.

1.51%​ Let X represent the random variable. The mean is μ=25,000, and the standard deviation is σ=6,000. The lower bound is negative infinity, which is represented by −9999, and the upper bound is 12,000. Now use a TI-83, TI-83 plus, or TI-84 calculator to find the probability. 1. Press 2nd and VARS for the DISTR menu. 2. Press 2 for normalcdf(. 3. Enter −9999,12000,25000,6000) and press ENTER. Then multiply the result by 100 to find a percentage. The answer, rounded to two decimal places, is 1.51%.

The Stomping Elephants volleyball team plays 30 matches in a week-long tournament. On average, they win 4 out of every 6 matches. What is the mean for the number of matches that they win in the tournament? 4/6 20 30 4

20 The mean of a binomial distribution is the product of p, the probability of a success, and n, the total number of repeated trials or events. mean = μ=np In this scenario, the number of trials is 30 (the total number of matches played), which is n. The probability of a success (winning a match) is 46. So, the mean of the binomial distribution B(30,46) is: μ=(30)(4/6)=20 .

A credit card company receives numerous phone calls throughout the day from customers reporting fraud and billing disputes. Most of these callers are put "on hold" until a company operator is free to help them. The company has determined that the length of time a caller is on hold is normally distributed with a mean of 2.5 minutes and a standard deviation of 0.5 minutes. If 2.5% of the callers are put on hold for longer than x minutes, what is the value of x? Use a TI-83, TI-83 plus, or TI-84 calculator. Round your answer to two decimal places.

3.48​ Here, the mean, μ, is 2.5 and the standard deviation, σ, is 0.5. The area to the right of x is 2.5%=0.025. The area to the left of x is 1−0.025=0.975. The value of x can be calculated using a TI-83, TI-83 plus, or TI-84 calculator. 1. Press 2nd and then VARS for the DISTR menu. 2. Press 3 for invNorm(. 3. Enter 0.975,2.5,0.5) and press ENTER. The answer, rounded to two decimal places, is x≈3.48. Thus, 2.5% of the callers are put on hold for longer than 3.48 minutes.

Natasha records the price for a gallon of home-heating oil from 10 randomly selected providers in her region on May 15. She does it again with another 10 randomly selected providers on November 15. The results (in dollars) are shown in the accompanying samples. Based on the z-scores you calculated above, would a price of $2.899 be more likely on May 15 or on November 15? A price of $2.899 would be more likely on May 15, because the absolute value of the z-score for this price on May 15 is less than on November 15. A price of $2.899 would be more likely on May 15, because the absolute value of the z-score for this price on May 15 is greater than on November 15. A price of $2.899 would be More likely on November 15, because the absolute value of the z-score for this price on May 15 is less than on November 15. A price of $2.899 would be more likely on November 15, because the absolute value of the z-score for this price on May 15 is greater than on November 15.

A price of $2.899 would be more likely on November 15, because the absolute value of the z-score for this price on May 15 is greater than on November 15. A z-score with a lower absolute value means that the data value is more likely to occur. Since the absolute value of the z-score for November 15, 1.74, is less than the absolute value of the z-score for May 15, 3.27, a price of $2.899 would be more likely on November 15.

A recent study was conducted to determine the concentration of lead in drinking water across various states. 25 water samples from the state of Washington were tested; the mean lead concentration was found to be 0.15%. This compares to 32 water samples in the state of Minnesota where the mean was reported to be 0.21%. The data was calculated to be significant at the 0.0001 level. Determine the meaning of this significance level. At the 0.0001 level of significance, the concentration of lead is higher in water from Minnesota. It is not unusual to see a 0.15% concentration of lead in drinking water because lead in drinking water is highly regulated. We can expect that any 32 water samples from Minnesota will have an average lead concentration of 0.21%. It is certain that drinking water in Minnesota has a higher lead concentration than drinking water in Washington.

At the 0.0001 level of significance, the concentration of lead is higher in water from Minnesota.

A recent survey was conducted to determine if political party has an impact on whether a person believes that higher education is necessary for career advancement. In the survey, 410 Republicans and 409 Democrats were surveyed. Of the Republicans, 36% believed that higher education was necessary for people to advance their career. This is very different from the 65% of Democrats surveyed who stated that higher education was necessary. The calculations were statistically significant at the 0.01 level of significance. What is the correct interpretation of the probability? We expect that 65% of all Democrats believe that higher education is necessary for career advancement. At the 0.01 level of significance, political party is associated with whether a person believes that higher education is necessary for career advancement. We cannot say that the results are statistically significant at the 0.05 level of significance. At the 0.01 level of significance, political party determines whether a person believes that higher education is necessary for career advancement.

At the 0.01 level of significance, political party is associated with whether a person believes that higher education is necessary for career advancement. The results of the poll are statistically significant at the 0.01 level of significance. This means that there is a 0.01 chance or less that the difference in opinion between Republicans and Democrats was the result of chance. Because of this, we can be fairly confident that the two parties have different opinions of higher education's necessity in career advancement.

A poll is conducted to determine if political party has any association with whether a person is for or against a certain bill. In the poll, 214 out of 432 Democrats and 246 out of 421 Republicans are in favor of the bill. Assuming political party has no association, the probability of these results being by chance is calculated to be 0.01. Interpret the results of the calculation. We can expect that 246 out of every 421 Republicans are in support of this bill. We cannot say the results are statistically significant at the 0.05 level. At the 0.01 level of significance, political party is associated with whether a person supports this bill. At the 0.01 level of significance, political party determines whether a person supports this bill.

At the 0.01 level of significance, political party is associated with whether a person supports this bill. The results of the survey are significant at the 0.01 level. This means the probability that the data were the result of chance is 0.01 or less. Because of this, we can be fairly confident, but not certain, that the difference between the proportion of Democrats who are in favor of the bill and the proportion of Republicans who are in favor of the bill is significant.

A group conducting a survey randomly selects adults in a certain region. Of the 2,500 adults selected, 1,684 are men. Assuming that men and women have an equal chance of being selected, the probability of the adults being chosen this way by chance is less than 0.01. Interpret the results of this calculation. We can expect that of any 2,500 adults selected for the survey, 1,684 of them will be men. We can be certain that the group is more likely to select men over women. At the 0.01 level of significance, the group is more likely to select men over women. The results are not statistically significant at the 0.05 level.

At the 0.01 level of significance, the group is more likely to select men over women. The results of the survey are significant at the 0.01 level. This means that the probability of the group's selection being the result of chance is 0.01 or less. Because of this, we can be fairly confident, but not certain, that the group is more likely to select men over women. FEEDBACK

The mean body temperature of a human is accepted to be 98.6∘F. In a study, the body temperatures of 127 individuals were measured. The mean body temperature of the individuals was calculated to be 99.0∘F. Assuming the regular body temperature of humans is actually 98.6∘F, the probability of these results occurring by chance is less than 0.01. Interpret the results of the calculation. We can expect that the mean body temperature of any 127 individuals is 99.0∘F. The mean body temperature of humans is not 98.6∘F. At the 0.01 level of significance, the mean body temperature of humans is not 98.6∘F. Since human body temperature varies, the results of this study are not unusual.

At the 0.01 level of significance, the mean body temperature of humans is not 98.6∘F. The results of the survey are significant at the 0.01 level. This means that the probability of the data being the result of chance is 0.01 or less. Because of this, we can be fairly confident, but not certain, that the mean body temperature of humans is not 98.6∘F.

A farmer claims that the average mass of an apple grown in his orchard is 100g. To test this claim, he measures the mass of 150 apples that are grown in his orchard and determines the average mass per apple to be 98g. The results are calculated to be statistically significant at the 0.01 level. What is the correct interpretation of this calculation? The data are not statistically significant at the 0.05 level. The mean mass of any 150 apples grown in the farmer's orchard is 98g. At the 0.01 level of significance, the mean mass of the apples grown in the farmer's orchard is different from 100g. At the 0.01 level of significance, the mean mass of the apples grown in the farmer's orchard is 98g.

At the 0.01 level of significance, the mean mass of the apples grown in the farmer's orchard is different from 100g. The results are statistically significant at the 0.01 level. This means the probability that the data the farmer gathered were due to chance is 0.01 or less. Therefore, the farmer can be fairly confident that the mean mass of any apple grown in his orchard is different from 100g.

One symptom of asthma is coughing. In a clinical trial of a new drug intended to treat asthma, 4 of the 90 subjects in the treatment group experienced a cough, and 12 of the 150 subjects in the control group had a cough. The difference is calculated to be significant at the 0.05 level. Determine the meaning of this significance level. We can expect an average of 4 out of every 90 subjects who receive the experimental drug to experience a cough. At the 0.05 level of significance, the new drug is effective at reducing asthma symptoms. It is not unusual to see 12 out of every 150 subjects cough because anyone can cough at any time. It is certain that this new drug is effective at treating asthma.

At the 0.05 level of significance, the new drug is effective at reducing asthma symptoms. The results of the test are significant at the 0.05 level. This means the probability that the outcome was the result of chance is 0.05 chance or less. Because of this, we can be fairly confident, but not certain, that the drug is effective at treating asthma.

Key Terms

Experiment: a planned operation carried out under controlled conditions. Trial: one repetition or instance of an experiment. Outcome: any possible result of an experiment. Sample space: all possible outcomes of an experiment. Event: a subset of the the sample space. Probability: a measure of a particular event happening compared to all events in a sample space. Likelihood: an interpretation of how likely an event is to occur.

A hospital takes record of any birth that occurs there every day. On one day, the hospital reports that 35 of the 62 babies born were girls. Assuming that all of the parents did not have any gender selection procedures, there is a probability of 0.31 of getting these results by chance. Do these results have statistical significance at the 0.05 level of significance? Yes, the probability of the results occurring is less than 0.05. No, the probability of the results occurring is less than 0.05. Yes, the probability of the results occurring is greater than 0.05. No, the probability of the results occurring is greater than 0.05.

No, the probability of the results occurring is greater than 0.05. Since the probability, 0.31, is greater than 0.05, the the results do not have statistical significance at the 0.05 level of significance.

For a certain animal, suppose that the number of babies born is independent for each pregnancy. This animal has a 70% chance of having 1 baby and a 30% chance of having 2 babies at each pregnancy. Let X be a random variable that represents the total number of babies if the animal gets pregnant twice. Construct a table showing the probability distribution of X. Arrange x in increasing order. Write the probabilities P(x) as decimals rounded to two decimals.

If the animal gets pregnant twice, the animal can have either one of the below combinations, where the first number represents the number of babies in the first pregnancy and the second number represents the number of babies in the second pregnancy. {1,1}{1,2}{2,1}{2,2} The probability of having1baby in the first pregnancy is70%=0.70and the probability of having1baby in the second pregnancy is70%=0.70, so the probability of{1,1}is0.70×0.70=0.49. The probability of having 1 baby in the first pregnancy is70%=0.70and the probability of having2babies in the second pregnancy is30%=0.30, so the probability of{1,2}is0.70×0.30=0.21. The probability of having2babies in the first pregnancy is30%=0.30and the probability of having1baby in the second pregnancy is70%=0.70, so the probability of{2,1}is0.30×0.70=0.21. The probability of having2babies in the first pregnancy is30%=0.30and the probability of having2babies in the second pregnancy is30%=0.30, so the probability of{2,2}is0.30×0.30=0.09. Here, X represents the total number of babies if the animal gets pregnant twice. Thus, x takes 2, 3, or 4. P(2)=probability of {1,1}=0.49, P(3)=probability of {1,2}+probability of {2,1}=0.21+0.21=0.42, and P(4)=probability of {2,2}=0.09. Thus, the probability distribution is shown below. x 2 3 4 P(x) 0.49 0.42 0.09

At a sports event, a fair coin is flipped to determine which team has possession of the ball to start. The coin has two sides, heads, (H), and tails, (T). Identify the correct experiment, trial, and outcome below: The experiment is identifying whether a heads or tails is flipped. The experiment is flipping the coin. A trial is flipping a heads. A trial is one flip of the coin. An outcome is flipping a tails. An outcome is flipping a coin once.

The experiment is flipping the coin. A trial is one flip of the coin. An outcome is flipping a tails.

In a large population, about 65% of families take a vacation during the summer. A researcher takes a random sample of 16 families and surveys whether they take a summer vacation. Use the binomial distribution to compute the probability that exactly 8 of the families take a summer vacation. To show your answer: Move the purple dots to identify the number of trials, n, and the probability of success, p. Use the black dot on the x-axis to represent the number of successes and display the probability. powered by Probability of 8 successes in 16 trials, S Number of Trials Probability of Success n = 16 p = 0.65 with a probability of success 0.65 is 0.0923

Since we are counting the number of families that take a vacation during the summer, and we know about 65% of all families take a vacation during the summer, the probability of success is p=0.65. Since there are 16 families in the sample, and each family is independent of others, we have n=16 independent trials. The problem asks for the probability that exactly 8 of the 16 families take a vacation during the summer. Using the binomial distribution, we set the purple dot sliders to p=0.65 and n=16, and we set the black dot slider to 8 to see the probability is 0.0923.

People with some college education stay married to their spouses for at least twenty years at a rate of 67%. How likely is it that a randomly selected, college educated couple will be married for at least twenty years? Very likely, the probability is close to 1. Somewhat likely, the probability is closer to 1 than to 0. Unlikely, the probability is close to 0. Somewhat unlikely, the probability is closer to 0 than it is to 1. Equally likely, the probability is 0.5.

Somewhat likely, the probability is closer to 1 than to 0.

Flights from Miami for New York City are delayed 27% of the time. How likely is it that a particular traveler's flight from Miami to New York will be delayed? Very likely, the probability is close to 1. Somewhat likely, the probability is closer to 1 than to 0. Unlikely, the probability is close to 0. Somewhat unlikely, the probability is closer to 0 than it is to 1. Equally likely, the probability is 0.5.

Somewhat unlikely, the probability is closer to 0 than it is to 1.

Flights from Miami for New York City are delayed 27% of the time. How likely is it that a particular traveler's flight from Miami to New York will be delayed? Very likely, the probability is close to 1. Somewhat likely, the probability is closer to 1 than to 0. Unlikely, the probability is close to 0. Somewhat unlikely, the probability is closer to 0 than it is to 1. Equally likely, the probability is 0.5.

Somewhat unlikely, the probability is closer to 0 than it is to 1. Here, we would be most correct to say that the probability is somewhat unlikely . It is not closer to 1 than it is to 0, so it is not appropriate to call it "likely" or "somewhat likely." It is not very close to 0, so it is not quite appropriate to call it "unlikely". It is false to call it "equally likely" since the probability isn't 0.5. We settle on "somewhat unlikely," since it is closer to 0 than to 1 but isn't that close to zero.

A university is trying to determine if certain remedial classes need to be offered to their incoming class. To do so, they give each student the same entrance exam and record their results. Any student scoring lower than a 75% will be recommended to a remedial course. Identify the correct experiment, trial, and outcome below: The experiment is identifying whether a student gets less than a 75% on the entrance exam. The experiment is giving the entrance exam. A trial is a student scoring less than 75% on an exam. The trial is one exam being taken. An outcome is a student scoring 80%. The outcome is the one student taking the entrance exam.

The experiment is giving the entrance exam. The trial is one exam being taken. An outcome is a student scoring 80%.

A credit card company requires that its customers choose a 3 digit, numerical code to access their online account. Each digit should be chosen from 0,1,2,3,4,5,6,7,8 or 9. So there are 10 possibilities for each digit. How many different three digit codes are there?

The Fundamental Counting Principle tells us that we multiple the number of choices for each digit by one another to get the total number of possibilities. Possibilities=10×10×10=1,000

A survey was conducted to see whether age has an association with the belief that a master's degree or higher provides an advantage in one's career. Of the 524 adults between the ages of 22 and 25 surveyed, 56% believed that a master's degree has value in a person's career path. Of the 458 adults surveyed between the ages of 40 and 45, 52% also believed that a master's degree has value in a person's career path. Assuming age is not associated with this belief, the probability of the data being the result of chance is calculated to be 0.21. Interpret this calculation. We can expect 56% of all adults between the ages of 22 and 25 to believe a master's degree or higher provides an advantage in one's career. The data is statistically significant at the 0.05 level of significance in showing that age has an association with the belief that a master's degree or higher provides an advantage in one's career. Age does not have any association with the belief that a master's degree or higher provides an advantage in one's career. The data are not statistically significant at the 0.05 level of significance in showing that age has an association with the belief that a master's degree or higher provides an advantage in one's career.

The data are not statistically significant at the 0.05 level of significance in showing that age has an association with the belief that a master's degree or higher provides an advantage in one's career. The probability calculated is 0.21. Since this is greater than 0.05, the data are not statistically significant in showing that age has an association with the belief that a master's degree or higher is advantageous in one's career path.

If the mean of a data set is 32 and the median is 36, which of the following is most likely? The data are skewed to the left. The data are skewed to the right. The data are symmetric.

The data are skewed to the left. Because the mean, 32, is less than the median, 36, we expect that there are some very small values which are bringing the mean down. In other words, the data are skewed to the left.

If the median of a data set is 13 and the mean is 23, which of the following is most likely? The data are skewed to the left. The data are skewed to the right. The data are symmetrical.

The data are skewed to the right. Because the mean, 23, is greater than the median, 13, we expect that there are some very large values which are bringing the mean up. In other words, the data are skewed to the right.

A poll was conducted to determine if there was any possible connection between men and women who live in a certain city and the favorability of a mayor in the city. In the poll, of the 400 men selected, 210 reported being in favor of the mayor. Of the 400 women selected, 190 also reported being in favor of the mayor. The probability of these results occurring by chance is calculated to be about 0.16. Interpret the results of the calculation at the 0.05 level of significance. It is expected that 210 of every 400 men in this city will be in favor of the mayor. The difference in the proportions is not statistically significant because the probability is greater than 0.05. Gender has no relation to the favorability of this mayor. The difference in the proportions shows that gender could have an association with how favorable the mayor is.

The difference in the proportions is not statistically significant because the probability is greater than 0.05. The difference in the proportions is significant at the 0.16 level. Since this value is greater than 0.05, we cannot say that gender has an association with favorability based on this poll.

A poll was conducted to determine if there was any possible connection between men and women who live in a certain city and the favorability of a mayor in the city. In the poll, of the 400 men selected, 210 reported being in favor of the mayor. Of the 400 women selected, 190 also reported being in favor of the mayor. The probability of these results occurring by chance is calculated to be about 0.16. Interpret the results of the calculation at the 0.05 level of significance. Select the correct answer below: It is expected that 210 of every 400 men in this city will be in favor of the mayor. The difference in the proportions is not statistically significant because the probability is greater than 0.05. Gender has no relation to the favorability of this mayor. The difference in the proportions shows that gender could have an association with how favorable the mayor is.

The difference in the proportions is not statistically significant because the probability is greater than 0.05. The difference in the proportions is significant at the 0.16 level. Since this value is greater than 0.05, we cannot say that gender has an association with favorability based on this poll.

A computer randomly generates numbers 1 through 100 for a lottery game. Every lottery ticket has 7 numbers on it. Identify the correct experiment, trial, and outcome below:

The experiment is the computer randomly generating a number. A trial is one number generated. An outcome is the number 2 being generated. An experiment is a planned operation carried out under controlled conditions. Here, generating a random number is the experiment. A trial is one instance of a an experiment taking place. A trial here is one number being generated. An outcome is any of the possible results of the experiment. Here, the outcome is any of the numbers between 1 and 100.

Determine whether or not the distribution is a probability distribution and select the reason(s) why or why not. x: 1, 2, 3 P(x): 1/4, 1/2, 1/4

The given distribution is a probability distribution, since the sum of probabilities is equal to 1. The given distribution is a probability distribution, since the probabilities lie inclusively between 0 and 1.

Determine whether or not the distribution is a probability distribution and select the reason(s) why or why not. X: 1, 2, 3, 4 P(X): 0.05, 1.05, 0.35, 0.55 The given distribution is a probability distribution, since the probabilities lie inclusively between 0 and 1. The given distribution is not a probability distribution, since the sum of probabilities is not equal to 1. The given distribution is a probability distribution, since the sum of probabilities is equal to 1. The given distribution is not a probability distribution, since at least one of the probabilities is greater than 1 or less than 0.

The given distribution is not a probability distribution, since the sum of probabilities is not equal to 1. The given distribution is not a probability distribution, since at least one of the probabilities is greater than 1 or less than 0.

A food processing plant fills snack-sized bags of crackers. The mean number of crackers in each bag is 22 and the standard deviation is 2. The factory supervisor selects one bag that contains 24 crackers. Which of the following statements is true? The number of crackers in the supervisor's bag is 2 standard deviations to the right of the mean. The number of crackers in the supervisor's bag is 2 standard deviations to the left of the mean. The number of crackers in the supervisor's bag is 1 standard deviation to the right of the mean. The number of crackers in the supervisor's bag is 1 standard deviation to the left of the mean.

The number of crackers in the supervisor's bag is 1 standard deviation to the right of the mean

Consider how the following scenario could be modeled with a binomial distribution, and answer the question that follows. 54.4% of tickets sold to a movie are sold with a popcorn coupon, and 45.6% are not. You want to calculate the probability of selling exactly 6 tickets with popcorn coupons out of 10 total tickets (or 6 successes in 10 trials). What value should you use for the parameter p?

The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a movie ticket with a popcorn coupon, so p=0.544.

Give the numerical value of the parameter p in the following binomial distribution scenario.A softball pitcher has a 0.692 probability of throwing a strike for each pitch and a 0.308 probability of throwing a ball. If the softball pitcher throws 29 pitches, we want to know the probability that more than 17 of them are strikes.Consider strikes as successes in the binomial distribution. Do not include p= in your answer.

The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.692.

Give the numerical value of the parameter p in the following binomial distribution scenario.A softball pitcher has a 0.721 probability of throwing a strike for each pitch and a 0.279 probability of throwing a ball. If the softball pitcher throws 19 pitches, we want to know the probability that more than 15 of them are strikes.Consider strikes as successes in the binomial distribution. Do not include p= in your answer.

The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.721.

Give the numerical value of the parameter p in the following binomial distribution scenario.The probability of winning an arcade game is 0.632 and the probability of losing is 0.368. If you play the arcade game 10 times, we want to know the probability of winning no more than 8 times.Consider winning as a success in the binomial distribution. Do not include p= in your answer.

The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is winning a game, so p=0.632.

Identify the parameters p and n in the following binomial distribution scenario. The probability of winning an arcade game is 0.718 and the probability of losing is 0.282. If you play the arcade game 20 times, we want to know the probability of winning more than 15 times. (Consider winning as a success in the binomial distribution.)

The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is winning a game, so p=0.718. The total number of trials, or games, is n=20.

Give the numerical value of the parameter n in the following binomial distribution scenario.The probability of buying a movie ticket with a popcorn coupon is 0.597 and without a popcorn coupon is 0.403. If you buy 18 movie tickets, we want to know the probability that no more than 13 of the tickets have popcorn coupons.Consider tickets with popcorn coupons as successes in the binomial distribution. Do not include n= in your answer.

The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, the total number of trials, or movie tickets, is n=18.

A city council sends out a survey to city residents to determine the support for a new school being built downtown. The survey asks residents whether they support the new school being built and their location in town. Out of those who respond, 376 of the 641 residents surveyed in Ward 1 support the new school being built, while 214 of the 398 residents surveyed in Ward 2 support the new school. Assuming the location of residents has no association with supporting the new school being built, the probability of the results of the survey being due to chance is calculated to be 0.12. Interpret the result of this calculation. The council can expect 376 of every 641 residents in Ward 1 to be in favor of building the new school. The results of the survey are statistically significant at the 0.05 level in showing that the location of city residents has an association with support for building the new school. The results of the survey are not statistically significant at the 0.05 level in showing that the location of the city's residents has an association with support for building the new school. The location of city residents is not associated with support for the new school.

The results of the survey are not statistically significant at the 0.05 level in showing that the location of the city's residents has an association with support for building the new school. The probability calculated is 0.12. Since the probability is greater than the level of significance, 0.05, the results are not statistically significant in showing that the location of city residents has an association with support for building the new school.

Three fair coins are flipped at the same time. Each coin has the two possible outcomes: heads or tails. There are 8 possible outcomes for the three coins being flipped: {HHH,TTT,HHT,HTT,THH,TTH,HTH,THT}. Let A be the event that the second coin flipped shows a head. Identify the numbers of each of the following:

There are 8 outcomes in the sample space. There are 4 outcomes in event A. P(A)=1/2​, is the probability that the second coin flipped shows heads.

Assume that the probability calculated above is 0.14. Give the best interpretation of that probability. 14% of families take a vacation during summer. 65% of families spend 14% of summer on vacation. There is a 14% probability that exactly half of the families in our sample take a summer vacation. The average family takes a vacation in summer 14% of the time.

There is a 14% probability that exactly half of the families in our sample take a summer vacation. The binomial probability that we calculated is the chance that exactly 8 of the 16 families in our sample take a summer vacation if 65% of all families take a summer vacation.

Assume the probability computed above is 0.08. Choose the best interpretation of this probability. 8% of people choose chocolate over vanilla. There is an 8% chance of selecting 3 people from the population who prefer chocolate over vanilla. There is an 8% chance of there being exactly 3 people who prefer chocolate over vanilla in our sample of 12 people. The average person prefers chocolate 8% of the time.

There is an 8% chance of there being exactly 3 people who prefer chocolate over vanilla in our sample of 12 people. The binomial probability that we calculated is the chance of exactly 3 people in our sample preferring chocolate over vanilla in our sample of 12 people.

Which of the following frequency tables shows a skewed data set? Select all answers that apply. Value Frequency74889121016111512131310145 ValueFrequency536378812915101911191210134143153161 ValueFrequency12113214315131610172618251915205 ValueFrequency09115218323421596372

ValueFrequency12113214315131610172618251915205 ValueFrequency09115218323421596372 Remember that data are left skewed if there is a main concentration of large values with several much smaller values. Similarly, right skewed data have a main concentration of small values with several much larger values. We can see that the following are left skewed because of the concentration of large values with many smaller values: ValueFrequency12113214315131610172618251915205 And the following is right skewed because of its concentration of small values with many larger values: ValueFrequency09115218323421596372 The other frequency table is more balanced and symmetrical.

A standard roulette wheel with slots labeled 0, 00, 1, 2, 3, ... , 36 is spun 60 times. Of these spins, the number 7 is spun 7 times. Assuming the wheel is fair, the probability of this occurring by chance is 0.00001. Do these results have statistical significance at the 0.01 level of significance? No, the probability of the results occurring is greater than 0.01. Yes, the probability of the results occurring is less than 0.01. No, the probability of the results occurring is less than 0.01. Yes, the probability of the results occurring is greater than 0.01.

Yes, the probability of the results occurring is less than 0.01. Since the probability of the results occurring, 0.00001, is less than 0.01, the data do have statistical significance at the 0.01 level.

Which of the following events seem like they would be unlikely to occur by chance? Select all that apply. flipping a coin 100 times and having it land on heads 15 times rolling a 6-sided die 100 times and having a 6 land face-up 15 times randomly selecting 10 people out of a crowd and having them all have brown eyes guessing somebody's name correctly without having previously met the person

flipping a coin 100 times and having it land on heads 15 times randomly selecting 10 people out of a crowd and having them all have brown eyes guessing somebody's name correctly without having previously met the person If the coin was fair, one would expect to see 50 heads. It would be very unlikely to observe only 15 heads by chance. While brown is the most common eye color, it would still be unlikely to pick 10 seemingly random people who all have brown eyes.Since there are many names, it seems like it would be unlikely to guess somebody's name correctly, even if one used a common name for one's guess.

In a large population, about 65% of people prefer chocolate ice cream over vanilla. A researcher takes a random sample of 12 people and surveys whether they prefer chocolate over vanilla ice cream. Use the binomial distribution to compute the probability that exactly 3 of the people in the sample prefer chocolate over vanilla ice cream. Identify the following information required to find the probability of people who prefer chocolate ice cream over vanilla.

n= 12 x= 3 p= 0.65 We determine n, the number of trials, by reviewing the context. The researcher randomly sampled 12 people. We determine x, the number of successes, by determining what results we are looking for. In this case, we'd like to know the probability for the number of times exactly 3 people in the sample prefer chocolate over vanilla ice cream. Finally, we determine the probability from the context, as we're told about 65% of people who prefer chocolate over vanilla ice cream.

In a large population, about 65% of families take a vacation during the summer. A researcher takes a random sample of 16 families and surveys whether they take a summer vacation. Use the binomial distribution to compute the probability that exactly 8 of the families take a summer vacation. Identify the following information required to find the probability of families that take a vacation during the summer.

n= 16 x= 8 p= 0.65 We determine n, the number of trials, by reviewing the context. The researcher randomly sampled 16 families. We determine x, the number of successes, by determining what results we are looking for. In this case, we'd like to know the probability for the number of times exactly 8 families take a vacation during the summer. Finally, we determine the probability from the context, as we're told about 65% of families take a vacation during the summer.

Identify the parameters p and n in the following binomial distribution scenario. The probability of winning an arcade game is 0.489 and the probability of losing is 0.511. If you play the arcade game 15 times, we want to know the probability of winning more than 8 times. (Consider winning as a success in the binomial distribution.)

n=15, p=0.489 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is winning a game, so p=0.489. The total number of trials, or games, is n=15.

Identify the parameters p and n in the following binomial distribution scenario. A basketball player has a 0.404 probability of making a free throw and a 0.596 probability of missing. If the player shoots 21 free throws, we want to know the probability that he makes no more than 6 of them. (Consider made free throws as successes in the binomial distribution.)

n=21, p=0.404 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a made free throw, so p=0.404. The total number of trials, or free throws, is n=21.

A dairy farmer milks his two cows every day. He determined the chance that he gets anywhere between 12 and 14 gallons of milk in one day is around 32%. Identify the method of probability the farmer used to reach this conclusion.

relative frequency There is no way the farmer could know before milking the cows how much milk he is going to get on any day. He would need to record the amount of milk gathered on previous days and estimate the probability of obtaining 12 to 14 gallons of milk. Thus, he used the relative frequency method.