CHAPTER 5 REVIEW

As a part of a promotion for a new type of cracker, free trial samples are offered to shoppers in a local supermarket. The probability that a shopper will buy a packet of crackers after tasting the free sample is 0.20. Different shoppers can be regarded as independent trials. If the random variable X is the number of the next 100 shoppers that buy a packet of the crackers after tasting a free sample, then the probability that 20 or more shoppers buy a packet is approximately... 1. The 100 trials are independent. 2. There is a fixed number of trials, 100. 3. There only two possible outcomes for each event. 4. The probability of success is fixed from trial to trial.

As a part of a promotion for a new type of cracker, free trial samples are offered to shoppers in a local supermarket. The probability that a shopper will buy a packet of crackers after tasting the free sample is 0.20. Different shoppers can be regarded as independent trials. If the random variable X is the number of the next 100 shoppers that buy a packet of the crackers after tasting a free sample, then the probability that 20 or more shoppers buy a packet is approximately... The random variable X is binomial. P(X  20) = 1- P(X  19) Using the binomial program on your TI or Excel I get approximately 0.5398. Excel: 1 - Binomdist(19, 100, .2, true) TI-83: 1 - Binomcdf(100, .2, 19)

There are 20 multiple choice questions on an exam, each having responses a, b, c, or d. Each question is worth 5 points and only one option per question is correct. Suppose the student guesses the answer to each question, and the guesses from question to question are independent. The probability that the student gets between 5 and 10 questions correct, including 5 and 10 is...

There are 20 multiple choice questions on an exam, each having responses a, b, c, or d. Each question is worth 5 points and only one option per question is correct. Suppose the student guesses the answer to each question, and the guesses from question to question are independent. The probability that the student gets between 5 and 10 questions correct, including 5 and 10 is... 1. The 20 trials are independent. 2. There is a fixed number of trials, 20. 3. There only two possible outcomes for each event. 4. The probability of success is fixed from trial to trial, p = 0.25.

There are 20 multiple choice questions on an exam, each having responses a, b, c, or d. Each question is worth 5 points and only one option per question is correct. Suppose the student guesses the answer to each question, and the guesses from question to question are independent. The probability that the student gets between 5 and 10 questions correct, including 5 and 10 is... The random variable X is binomial: n = 20 and p = 0.25 P(5  X  10) = P(X  10) - P(X  4)  58.12%. Excel: Binomdist(10,20,.25,true) - Binomdist(4,20, .25,true) TI 83: binomcdf(20,.25,10)-binomcdf(20,.25,4)

There are 20 multiple choice questions on an exam, each having responses a, b, c, or d. Each question is worth 5 points and only one option per question is correct. Suppose the student guesses the answer to each question, and the guesses from question to question are independent. The probability that the student gets between 5 and 10 questions correct, including 5 and 10 is... Would this problem be suitable to use the normal approximation? Since np = 20(0.25)  10 it would not be suitable to use a normal approximation.

An important result in statistics says that in many situations for large sample sizes the sampling distribution of the sample mean is approximately normal. This famous result is known as... a. the law of large numbers. b. the multiplication rule. c. the central limit theorem. d. the bell curve.

Suppose that you are a student worker in the statistics department and they agree to pay you using the random pay system. Each week the department chair flips a coin. If it comes up heads, your pay for the week is $80 and if it comes up tails your pay for the week is $40. Your friend is working for the engineering department and makes $65 per week. The probability that your total earnings in 100 weeks are on average more that your friends is approximately First we need to determine what type of distribution we have here. I will organize the information. The pay is determined by the flip of a coin. Since the variable of interest is the amount of money I have earned, let the random variable X, be the amount of money I have earned.

Suppose that you are a student worker in the statistics department and they agree to pay you using the random pay system. Each week the department chair flips a coin. If it comes up heads, your pay for the week is $80 and if it comes up tails your pay for the week is $40. Your friend is working for the engineering department and makes $65 per week. The probability that your total earnings in 100 weeks are on average more that your friends is approximately... X $40 $80 P(x) 0.5 0.5 Probability Distribution: When I look at the question I can see that I want to calculate P(X  $65). Thus, I am really considering the sampling distribution of X, for n = 100. In other words, for every run of 100, what percentage generate an average greater than $65?

Suppose that you are a student worker in the statistics department and they agree to pay you using the random pay system. Each week the department chair flips a coin. If it comes up heads, your pay for the week is $80 and if it comes up tails your pay for the week is $40. Your friend is working for the engineering department and makes $65 per week. The probability that your total earnings in 100 weeks are on average more that your friends is approximately... X $40 $80 P(x) 0.5 0.5 Probability Distribution: P(X  $65). Now I need to consider how to calculate this probability. The distribution X is not normal, it is discrete. However, the central limit theorem says that the sampling distribution of X is approximately normal for n large enough.

Suppose that you are a student worker in the statistics department and they agree to pay you using the random pay system. Each week the department chair flips a coin. If it comes up heads, your pay for the week is $80 and if it comes up tails your pay for the week is $40. Your friend is working for the engineering department and makes $65 per week. The probability that your total earnings in 100 weeks are on average more that your friends is approximately... X $40 $80 P(x) 0.5 0.5 Probability Distribution: P(X  $65). Thus, we need to find  and .  = 40(0.5) + 80 (0.5) = 20 + 40 = 60. 2 = (40-60)2 (0.5) + (80-60)2(0.5)  = 20

Suppose that you are a student worker in the statistics department and they agree to pay you using the random pay system. Each week the department chair flips a coin. If it comes up heads, your pay for the week is $80 and if it comes up tails your pay for the week is $40. Your friend is working for the engineering department and makes $65 per week. The probability that your total earnings in 100 weeks are on average more that your friends is approximately... X $40 $80 P(x) 0.5 0.5 Probability Distribution:  = 60 and  = 20 P(X  $65)  P (Z > ) P(Z > 2.5) = .0062

There are 20 multiple choice questions on an exam, each having responses a, b, c, or d. Each question is worth 5 points and only one option per question is correct. Suppose the student guesses the answer to each question, and the guesses from question to question are independent. The probability that the student gets no questions correct is... Let the random variable X count the number of incorrect responses. We note this is a binomial setting, but since we want to calculate an extreme event we can rely on the same technique we used in chapter 4. We want to calculate P(X = 20) = P(answer 1st question incorrect and second incorrect and ... and last incorrect)= (3/4)(3/4)...(3/4) = (3/4)20 = 0.00317

The weights of medium oranges packaged by an orchard are normally distributed with a mean of 14 ounces and a standard deviation of 2 ounces. What is the probability of choosing 4 oranges at random and having the average weight exceed 15 ounces? We want to calculate P(X > 15) where the random variable X is the average weight of 15 medium oranges. So this is a problem involving the sampling distribution of the mean. Why is it that I can change the random variable X into a Z score? P(X > 15) = P(Z > ) = P(Z > 1) = 0.1587

The weights of medium oranges packaged by an orchard are normally distributed with a mean of 14 ounces and a standard deviation of 2 ounces. What is the probability of choosing 4 oranges at random and having the average weight exceed 15 ounces? We want to calculate P(X > 15) where the random variable X is the average weight of 15 medium oranges. So this is a problem involving the sampling distribution of the mean. Why is it that I can change the random variable X into a Z score? The distribution is normal to begin with so the sampling distribution is also normally distributed.

A population’s distribution is normal A population’s distribution is normal. If we look at the sampling distribution of the mean for this population, where the sample size is n = 2, then we would find that a. the sampling distribution of the mean is not normal. b. the sampling distribution of the mean is approximately normal. c. the sampling distribution of the mean is exactly normal. d. it is impossible to know without running some simulations.

A population’s distribution is not normal A population’s distribution is not normal. If we look at the sampling distribution of the mean for this population, where the sample size is n = 40, then we would find that a. the sampling distribution of the mean is not normal. b. the sampling distribution of the mean is approximately normal. c. the sampling distribution of the mean is exactly normal. d. it is impossible to know without running some simulations.

A survey asks a random sample of 1500 adults in Ohio if they support an increase in the state sales tax from 5% to 6%, with the additional revenue going to education. Let p denote the proportion in the sample that say they support the increase. Suppose that 40% of all adults in Ohio support the increase. The probability that p is more than 0.45 is Instead of dealing with the population proportion, let us look a the count instead. If we surveyed 1500 adults in this population where the population proportion is 0.40 then, 0.40(1500) = 600, we would expect around 600 adults to be in favor of the proposal. So we want to know, .45(1500), P(X > 675), where the random variable X counts the number of people in favor of the proposal.

A survey asks a random sample of 1500 adults in Ohio if they support an increase in the state sales tax from 5% to 6%, with the additional revenue going to education. Let p denote the proportion in the sample that say they support the increase. Suppose that 40% of all adults in Ohio support the increase. The probability that p is more than 0.45 is So we want to know P(X > 675), where the random variable X counts the number of people in favor of the proposal. We can use a normal approximation here since np > 10 and n(1 - p) > 10 also. We have  = 600 and  = 18.97. P(X > 675)  P(Z > ) < .0002

The amount of coffee (in ounces) filled in a jar by a machine has a normal distribution with a mean amount of 17 ounces and a standard deviation of 0.2 ounces. A. The company producing this product can only sell jars that have 16 or more ounces of coffee. What percentage of the jars, in the long run, will not be sellable?

The amount of coffee (in ounces) filled in a jar by a machine has a normal distribution with a mean amount of 17 ounces and a standard deviation of 0.4 ounces. A. The company producing this product can only sell jars that have 16 or more ounces of coffee. What percentage of the jars, in the long run, will not be sellable? P(X < 16), where the random variable X is the amount of coffee in the jar in ounces. P(X < 16) = P(Z < ) = P(Z < -2.5) = 0.0062

The amount of coffee (in ounces) filled in a jar by a machine has a normal distribution with a mean amount of 17 ounces and a standard deviation of 0.4 ounces. B. If one were to randomly select 4 jars from the production line, what is the probability of 2 or more jars being rejected? Let the random variable X count the number of jars out of 4 that will be rejected. From the past problem we know that the probability of producing a rejected jar is 0.0062. Note: if we assume independence then we have a binomial situation. So, I will calculate P(X  2) using a binomial function. P(X  2) = 0.000229

The amount of coffee (in ounces) filled in a jar by a machine has a normal distribution with a mean amount of 17 ounces and a standard deviation of 0.4 ounces. C. If one were to randomly select 4 jars from the production line, what is the probability that the average amount of coffee in the four jars is 16.5 ounces or less? Since we are considering the average of the four jars, it is a sampling distribution problem. We want P(X < 16.5). So, I will calculate P(X < 16.5) = P(Z < ) = P(Z < -2.5) = 0.0062.

? THE END