Chapter 4 Review Questions Continuous or Discrete? 1. The number of students in a classroom. DISCRETE – students are counted. 2. The total IQ of the students in a classroom. CONTINUOUS – IQs are measured. 3. The total amount of rainfall in a year, in inches. CONTINUOUS – inches of rain are measured. 4. The number of days per year that it rains. DISCRETE – days of rain are counted. This is a probability distribution. (True or False?) 5. TRUE – all probabilities are between 0 and 1, and the sum is 1

Chapter 4 Review Questions This is a probability distribution. (True or False?) 6. FALSE – the sum is less than 1 (.888) 7. FALSE – not all probabilities are between 0 and 1. 8. FALSE – the sum is less than 1 (.9986) AND not all probabilities are between 0 and 1.

16.383 x Freq P(x) x*P(x) 𝒙−𝝁 (𝒙−𝝁 ) 𝟐 P(x) (𝒙−𝝁 ) 𝟐 15 18 .150 2.250 9. The mean of this distribution is: a) Find the probabilities for each x-value by dividing the frequencies by the total (120). b) Multiply each x-value by its probability. c) The sum of the x * P(x) column is the mean. d) The mean is 16.383 x Freq P(x) x*P(x) 𝒙−𝝁 (𝒙−𝝁 ) 𝟐 P(x) (𝒙−𝝁 ) 𝟐 15 18 .150 2.250 16 44 .367 5.867 17 52 .433 7.367 6 .050 .900 16.383

16.383 .636 x Freq P(x) x*P(x) 𝒙−𝝁 (𝒙−𝝁 ) 𝟐 P(x) (𝒙−𝝁 ) 𝟐 15 18 .150 10. The variance of this distribution is: a) subtract the mean from each x-value. b) square the answers to step a) to make them positive. c) multiply the values for P(x) by the values from step b). d) the sum of this last column is your variance. e) the variance is .636 if you wait to round until the end. if you rounded as you went, the variance is .637. 11. The standard deviation of this distribution is: a) the square root of the variance, either .797 or .798 x Freq P(x) x*P(x) 𝒙−𝝁 (𝒙−𝝁 ) 𝟐 P(x) (𝒙−𝝁 ) 𝟐 15 18 .150 2.250 -1.383 1.914 .287 16 44 .367 5.867 -0.383 .147 .054 17 52 .433 7.367 .617 .380 .165 6 .050 .900 1.617 2.614 .131 16.383 .636

CALCULATOR STEPS!! 9. The mean of this distribution is: STAT – Edit Put x-values into L1 Put frequencies into L2 Find the sum of L2 (2nd mode, 2nd Stat, Math 5, 2nd 2) Highlight L3 and enter L2/sum(L2) This puts the values for P(x) into L3 Highlight L2 and enter L3 This copies the P(x) values into L2, where you need them. Highlight L3 again, and enter L1*L2. Find the sum of L3 This is your mean. Your mean is 16.383

CALCULATOR STEPS!! 10. The variance of this distribution is: Highlight L4 and enter L1-sum(L3) Highlight L5 and enter L42 Highlight L6 and enter L2*L5 Find the sum of L6; this is your variance. Your variance is .636 11. The standard deviation of this distribution is: The standard deviation is the square root of your variance. .636 = .798

In a recent study of Probability and Statistics students at King's Fork High School, it was found that 92.5% of the students had an average above D. Consider a sample of 50 students. 12. What is the mean of this sample? μ=𝑛𝑝;𝜇=50 .925 =46.25 13. What is the variance of this sample? 𝜎 2 =𝑛𝑝𝑞=46.25∗.075=3.469 14. What is the standard deviation of this sample? σ= 𝑛𝑝𝑞 = 3.469 =1.862

It has been determined that 1 in five American adults lie on their income tax returns (made up number). In a random sample of 2350 American adults, what is the probability that: This is a BINOMIAL distribution. We know how many trials (2350) before we start Trials are independent, with the same probability (1/5 = .2) They either lied or they didn’t It would help to find the mean as a reference point: 2350 * .2 = 470. 15. exactly 500 of them lied on their tax return that year? 2nd VARS A (binompdf) (2350, .2, 500) = .0062 Use pdf because we want probability of one outcome.

It has been determined that 1 in five American adults lie on their income tax returns (made up number). In a random sample of 2350 American adults, what is the probability that: 16. less than 500 of them lied on their tax return that year? This is a cumulative probability, so we use binomcdf 2nd VARS B (binomcdf) (2350, .2, 499) = .9351 499 is the largest number less than 500, so that’s what we use. This answer makes sense since we are starting above the mean and going down past the mean all the way to zero. 17. more than 500 of them lied on their tax return that year? We need to use the complement rule here; calculator only gives probabilities for less than or equal to. 1 - 2nd VARS B (binomcdf) (2350, .2, 500) = .0587 500 is the smallest number not more than 500, so that’s what we use. This answer makes sense since we are starting above the mean and going higher, even further away from the expected value.

Last year, a certain football wide receiver caught at least one touchdown pass in 14 out of 16 games. If the player maintains that same rate, what is the probability that his first game with at least one touchdown reception this year happens: This is a GEOMETRIC distribution. We want to know when the first success will happen. Trials are independent 18. in the second game? We will use geometpdf, since we want the probability for one particular game (game 2) p is 14/16, since that’s the rate of success from last year. 2nd VARS E (geometpdf) (14/16, 2) = .109

Last year, a certain football wide receiver caught at least one touchdown pass in 14 out of 16 games. If the player maintains that same rate, what is the probability that his first game with at least one touchdown reception this year happens: 19. before the third game? This will be a geometcdf, since we want the cumulative probability of a success in either game 1 or game 2. 2nd Vars F (geometcdf) (14/16, 2) = .984 We use 2 for x because that’s the last game before the third game. 20. after the second game? This is the complement to #19. 1 - .984 = .016

If the number of hits on a business's web site is 12 per minute, what is the probability that there are: This is a Poisson distribution, since we are looking for a specific number of successes within a given unit of time, space or volume. 21. exactly 15 hits during any randomly selected minute? Use Poissonpdf to find the probability of exactly one x-value 2nd VARS C (12, 15) = .072 22. more than 13 hits during any randomly selected minute? We should know that this will be less than 50%, since we are starting above the average and going even further up from there. This is a Poissoncdf distribution; we want a cumulative probability. It’s also a complement rule question. 1 - 2nd VARS D (12,12) = .424

If the number of hits on a business's web site is 12 per minute, what is the probability that there are: 23. less than 15 hits during any randomly selected minute? We should know that this will be greater than 50%, since we are starting above the average and going down across the average to zero. This is a Poissoncdf distribution; we want a cumulative probability. This is NOT a complement rule question; we want less than, and that’s what the calculator will give us. 2nd VARS D (12,14) = .772 We use 14 for x since that’s the largest number less than 15.

A ball pit at the local McDonald's has 677 red balls, 876 blue balls, 712 green balls and 498 gold balls. If you randomly pull balls out, record their color and then replace them, what is the probability that the first gold ball you pull out happens: This is a geometric distribution; we are looking for our first success. Trials are independent because we are putting the balls back each time. Add up the number of balls of each color to find the total number of balls (2763) p = 498/2763 24. on the fifth try? 2nd VARS E (498/2763, 5) = .081

A ball pit at the local McDonald's has 677 red balls, 876 blue balls, 712 green balls and 498 gold balls. If you randomly pull balls out, record their color and then replace them, what is the probability that the first gold ball you pull out happens: 25. after the third try? This is complement rule, since we want more than 3. We will use 3 for x since that’s the largest number not more than itself. 1 – geometcdf (498/2763, 3) = .551 26. before the third try? geometcdf (498/2763, 2) = .328

A researcher observes cars as they go by and records their colors A researcher observes cars as they go by and records their colors. She observes 1458 cars, of which 256 are yellow. What is the probability that if she observes another 2500 cars, another binomial distribution. She knows that she will repeat the experiment 2500 times. The probability of each car being yellow is the same (256/1458). The car is either yellow or it isn’t. Wouldn’t hurt to find the mean (expected value) for comparison. (256/1458)*2500 = 438.96; we would expect about 439 yellow cars. 27. at least 440 of them will be yellow? This means 440 or more (use the complement rule). Since the expected value is 439, this will be close to .5, but less than .5 1 – binomcdf (2500, 256/1458, 439) = .486 .486 is close to .5, but less than .5; our answer makes sense.

A researcher observes cars as they go by and records their colors A researcher observes cars as they go by and records their colors. She observes 1458 cars, of which 256 are yellow. What is the probability that if she observes another 2500 cars, 28. at most 440 of them will be yellow? This means 440 or less. Since the expected value is 439, this will be close to .5, but greater than .5. 2nd VARS B (2500, 256/1458, 440) = .535 Notice that #28 and #27 are NOT complementary! The probability of exactly 440 is included in both questions.

A researcher observes cars as they go by and records their colors A researcher observes cars as they go by and records their colors. She observes 1458 cars, of which 256 are yellow. What is the probability that if she observes another 2500 cars, 29. between 430 and 445 (inclusive) will be yellow? We need to use binomcdf for up to 445 and for below 430 and then subtract the answers to find the area between them. binomcdf(2500, 256/1458, 445) – binomcdf(2500, 256/1458, 429) = .325 This makes sense since the average (expected value) is close to halfway between 430 and 445, so we would expect a fairly high probability.