Probability & Statistics Chapter 4 Test Review

Decide whether the random variable is discrete or continuous. X represents the number of motorcycle accidents in one year in California. Discrete X represents the volume of blood drawn for a blood test. Continuous

A company gave psychological tests to prospective employees A company gave psychological tests to prospective employees. The random variable represents the possible test scores. Use the histogram to find the probability that a person selected at random from the survey’s sample had a test score of (a) more than two and (b) less than four. 𝑷 𝒕𝒆𝒔𝒕 𝒔𝒄𝒐𝒓𝒆 >𝟐 =𝟎.𝟑𝟓 𝑷 𝒕𝒆𝒔𝒕 𝒔𝒄𝒐𝒓𝒆<𝟒 =𝟎.𝟗𝟎

Determine the probability distribution’s missing probability value. A sociologist surveyed the households in a small town. The random variable represents the number of dependent children in the households. 𝑷 𝟑 𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 𝒄𝒉𝒊𝒍𝒅𝒓𝒆𝒏 =𝟎.𝟐𝟐

Use the frequency distribution to construct a probability distribution, find the mean, variance, and standard deviation of the probability distribution. Round all numbers to the nearest third decimal place except for the standard deviation (nearest tenth). The following is a probability distribution of the number of dogs found in the respective households in a small town. Complete the table and missing values. 𝐝𝐨𝐠𝐬,𝒙 households 𝑷(𝒙) 𝒙𝑷(𝒙) 𝒙−𝝁 𝒙−𝒖 𝟐 𝒙−𝝁 𝟐 𝑷(𝒙) 1466 1 582 2 329 3 165 4 96 Frequency = Variance = 𝑃 𝑥 = Standard deviation = Mean = Expected value =

Use the frequency distribution to construct a probability distribution, find the mean, variance, and standard deviation of the probability distribution. Round all numbers to the nearest third decimal place except for the standard deviation (nearest tenth). The following is a probability distribution of the number of dogs found in the respective households in a small town. Complete the table and missing values. 𝐝𝐨𝐠𝐬,𝒙 households 𝑷(𝒙) 𝒙𝑷(𝒙) 𝒙−𝝁 𝒙−𝒖 𝟐 𝒙−𝝁 𝟐 𝑷(𝒙) 1466 0.556 0.000 −0.804 0.646 0.359 1 582 0.221 0.196 0.038 0.008 2 329 0.125 0.250 1.196 1.430 0.179 3 165 0.063 0.189 2.196 4.822 0.304 4 96 0.036 0.144 3.196 10.214 0.368 Frequency = 2638 Variance = 1.218 𝑃 𝑥 =𝟏.𝟎𝟎𝟏≈𝟏 Standard deviation = 1.1 Mean = 0.804 Expected value = 0.804

I Independent Each trial is independent of the other trials. A binomial experiment is a probability experiment that satisfies the following conditions. B Binary There are only two possible outcomes of interest for each trial. The outcomes can be classified as a success or as a failure. I Independent Each trial is independent of the other trials. N Number The experiment is repeated for a fixed number of trials S Success The probability of a success is the same for each trial.

𝑃 𝑥 = nCx 𝑝 𝑥 𝑞 𝑛−𝑥 = 𝑛! 𝑛−𝑥 !(𝑥!) 𝑝 𝑥 𝑞 𝑛−𝑥 You are taking a multiple-choice quiz that consists of five questions. Each question has four possible answers, only one of which is correct. To complete the quiz, you randomly guess the answer to each question. Find the probability of guessing (a) exactly three answers correctly,(b) at least three answers correctly, and (c) less than three answers correctly. Use the formula to get your answers. 𝑛=5 𝑝=0.25 𝑞=0.75 𝑥=0, 1, 2, 3, 4, 5 𝑃 𝑥 = nCx 𝑝 𝑥 𝑞 𝑛−𝑥 = 𝑛! 𝑛−𝑥 !(𝑥!) 𝑝 𝑥 𝑞 𝑛−𝑥 𝑃 𝑥=0 = 5C0 0.25 0 (0.75) 5 = 5! 5−0 !(0!) 0.25 0 ( 0.75) 5 ≈𝟎.𝟐𝟑𝟕 𝑃 𝑥=1 = 5C1 0.25 1 (0.75) 4 = 5! 5−1 ! 1! 0.25 1 ( 0.75) 4 ≈𝟎.𝟑𝟗𝟔 𝑃 𝑥=2 = 5C2 0.25 2 (0.75) 3 = 5! 5−2 ! 2! 0.25 2 ( 0.75) 3 ≈𝟎.𝟐𝟔𝟒 𝑃 𝑥=3 = 5C3 0.25 3 (0.75) 2 = 5! 5−3 ! 3! 0.25 3 ( 0.75) 2 ≈𝟎.𝟎𝟖𝟖

𝑃 𝑥=0 = 5C0 0.25 0 (0.75) 5 = 5! 5−0 !(0!) 0.25 0 ( 0.75) 5 ≈𝟎.𝟐𝟑𝟕 𝑃 𝑥=1 = 5C1 0.25 1 (0.75) 4 = 5! 5−1 ! 1! 0.25 1 ( 0.75) 4 ≈𝟎.𝟑𝟗𝟔 𝑃 𝑥=2 = 5C2 0.25 2 (0.75) 3 = 5! 5−2 ! 2! 0.25 2 ( 0.75) 3 ≈𝟎.𝟐𝟔𝟒 𝑃 𝑥=3 = 5C3 0.25 3 (0.75) 2 = 5! 5−3 ! 3! 0.25 3 ( 0.75) 2 ≈𝟎.𝟎𝟖𝟖 𝑷 𝒆𝒙𝒂𝒄𝒕𝒍𝒚 𝒕𝒉𝒓𝒆𝒆 𝒂𝒏𝒔𝒘𝒆𝒓𝒔 𝒄𝒐𝒓𝒓𝒆𝒄𝒕𝒍𝒚 =𝟎.𝟎𝟖𝟖 𝑷(less than three answers correctly)=𝑷 𝟎 +𝑷 𝟏 +𝑷 𝟐 =𝟎.𝟐𝟑𝟕+𝟎.𝟑𝟗𝟔+𝟎.𝟐𝟔𝟒=𝟎.𝟖𝟗𝟕 𝑷 𝒂𝒕 𝒍𝒆𝒂𝒔𝒕 𝒕𝒉𝒓𝒆𝒆 𝒂𝒏𝒔𝒘𝒆𝒓𝒔 𝒄𝒐𝒓𝒓𝒆𝒄𝒕𝒍𝒚 =𝟏−𝟎.𝟖𝟗𝟕=𝟎.𝟏𝟎𝟑

𝑷 𝒂𝒕 𝒎𝒐𝒔𝒕 𝟓 𝒘𝒐𝒓𝒌𝒆𝒓𝒔 =𝑷 𝟎 +𝑷 𝟏 +𝑷 𝟐 +𝑷 𝟑 +𝑷 𝟒 +𝑷 𝟓 Fourteen percent of workers believe they will need less than $250,000 when they retire. You randomly select 10 workers and ask each how much money he or she thinks they will need for retirement. Find the probability that the number of workers who say they will need less than $250,000 when they retire is (a) exactly two, and (b) at most five. Use the graphing calculator to get your answers. 𝑃 𝑥=0 =0.221 𝑃 𝑥=3 =0.115 𝑃 𝑥=1 =0.360 𝑃 𝑥=4 =0.033 𝑃 𝑥=2 =0.264 𝑃 𝑥=5 =0.006 𝑷 𝒆𝒙𝒂𝒄𝒕𝒍𝒚 𝒕𝒘𝒐 𝒘𝒐𝒓𝒌𝒆𝒓𝒔 =𝟎.𝟐𝟔𝟒 𝑷 𝒂𝒕 𝒎𝒐𝒔𝒕 𝟓 𝒘𝒐𝒓𝒌𝒆𝒓𝒔 =𝑷 𝟎 +𝑷 𝟏 +𝑷 𝟐 +𝑷 𝟑 +𝑷 𝟒 +𝑷 𝟓 =(𝟎.𝟐𝟐𝟏+𝟎.𝟑𝟔𝟎+𝟎.𝟐𝟔𝟒+𝟎.𝟏𝟏𝟓+𝟎.𝟎𝟑𝟑+𝟎.𝟎𝟎𝟔) =𝟎.𝟗𝟗𝟗

Thirty-seven percent of women consider themselves fans of professional baseball. You randomly select six women and ask each if she considers herself a fan of professional baseball. Construct a binomial distribution Graph the binomial distribution using a histogram Find the mean, variance, and standard deviation of the binomial distribution. 𝑛=6 𝑝=0.37 𝑞=0.63 𝑥=0, 1, 2, 3, 4, 5, 6 Use technology to help construct a binomial distribution (table).

Scroll down to binompdf Trials = 6 p=0.37 Go to 2nd Distr. Scroll down to binompdf Trials = 6 p=0.37 X−Value=0, then 1, then 2, then 3, then 4, then 5, then 6 Paste to the home screen, then Enter. Construct a binomial distribution 𝒙 𝑷(𝒙) 1 2 3 4 5 6

Scroll down to binompdf Trials = 6 p=0.37 Go to 2nd Distr. Scroll down to binompdf Trials = 6 p=0.37 X−Value=0, then 1, then 2, then 3, then 4, then 5, then 6 Paste to the home screen, then Enter 𝒙 𝑷(𝒙) 0.063 1 0.220 2 0.323 3 0.253 4 0.112 5 0.026 6 0.003 Are all the probabilities between 0 and 1? yes Add up the probabilities. 𝒑 𝒙 =𝟏

Graph the binomial distribution using a histogram The random variable x is on the x-axis. Place the numbers below the middle of each bar. Label both x-axis and y-axis. The y-axis is probability. Title the graph Use increments of 0.05 on the y-axis. Make sure the bars touch. The bars must be of equal width.

Find the mean, variance, and standard deviation of the binomial distribution. Round the mean and variance to three decimal places, and round the standard deviation to the nearest tenth. We need to know the values for 𝑛, 𝑝, 𝑞. 𝑛=6 𝑝=0.37 𝑞=0.63 𝑚𝑒𝑎𝑛=𝜇=𝑛𝑝= 6 0.37 =𝟐.𝟐𝟐 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒= 𝜎 2 =𝑛𝑝𝑞= 6 0.37 0.63 ≈𝟏.𝟑𝟗𝟗 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛=𝜎= 𝑛𝑝𝑞 ≈𝟏.𝟐

This is a geometric probability and we use the formula 𝑝 𝑥 =𝑝 (𝑞) 𝑥−1 Basketball player Shaquille O’Neal makes a free-throw shot about 52.6% of the time. Find the probability that (a) the first shot O’Neal makes is the second shot, (b) the first shot O’Neal makes is the first or second shot. This is a geometric probability and we use the formula 𝑝 𝑥 =𝑝 (𝑞) 𝑥−1 𝑝=0.526 𝑞=0.474 𝒑 𝒔𝒆𝒄𝒐𝒏𝒅 𝒔𝒉𝒐𝒕 = 𝟎.𝟓𝟐𝟔 𝟎.𝟒𝟕𝟒 𝟐−𝟏 = 𝟎.𝟓𝟐𝟔 𝟎.𝟒𝟕𝟒 𝟏 = 𝟎.𝟓𝟐𝟔 .𝟒𝟕𝟒 =𝟎.249 𝑝 𝑓𝑖𝑟𝑠𝑡 𝑜𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 𝑠ℎ𝑜𝑡 =𝑝 𝑓𝑖𝑟𝑠𝑡 𝑠ℎ𝑜𝑡 +𝑝(𝑠𝑒𝑐𝑜𝑛𝑑 𝑠ℎ𝑜𝑡) 𝑝 𝑠𝑒𝑐𝑜𝑛𝑑 𝑠ℎ𝑜𝑡 ≈0.249 𝑝 𝑓𝑖𝑟𝑠𝑡 𝑠ℎ𝑜𝑡 = 0.526 0.474 1−1 = 0.526 0.474 0 = 0.526 1 =0.526 𝒑 𝒇𝒊𝒓𝒔𝒕 𝒐𝒓 𝒔𝒆𝒄𝒐𝒏𝒅 𝒔𝒉𝒐𝒕 =𝟎.𝟓𝟐𝟔+𝟎.𝟐𝟒𝟗=𝟎.𝟕𝟕𝟓