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COB 191 > Exam 2 > Flashcards

Exam 2 Flashcards

1
Q

A lab orders 100 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Prices for 100 rats follow the following distribution:

price: 10.00 12.50 15.00
probability: 0.35 0.40 0.25

How much should the lab budget for next year’s rat orders, assuming this distribution does not change?

A

$637

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2
Q

In a binomial distribution

the random variable X is continuous.

the probability of success p is stable from trial to trial.

the number of trials, n, must be at least 30.

the results of one trial are dependent on the results of the other trials.

A

the probability of success p is stable from trial to trial.

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3
Q

The connotation “expected value” or “expected gain” from playing roulette at a casino means

the amount you expect to “gain” on a single play.

the amount you expect to “gain” in the long run over many plays.

the amount you need to “break even” over many plays.

the amount you should expect to gain if you are lucky.

A

the amount you expect to “gain” in the long run over many plays.

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4
Q

A fair coin is flipped 6 times. What is the probability that it comes up heads in exactly 4 out of the 6 times?
=BINOMDIST(6, 4, 2/3, TRUE)
=BINOMDIST(4, 6, 0.5, TRUE)
=BINOMDIST(4, 6, 0.5, FALSE)
=BINOMDIST(4, 6, 2/3, TRUE)
=BINOMDIST(6, 4, 2/3, FALSE)

A

=BINOMDIST(4, 6, 0.5, FALSE)

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5
Q

Think about the meaning of the quantity represented by the Excel expression =BINOMDIST(3, 27, 0.1, FALSE). Logically, which of the following must be equal to this quantity?
=BINOMDIST(24, 27, 0.1, FALSE)
=BINOMDIST(24, 27, 0.9, FALSE)
=1-BINOMDIST(24, 27, 0.1, FALSE)
=1-BINOMDIST(24, 27, 0.9, FALSE)
=BINOMDIST( 3, 24, 0.1, FALSE)

A

=BINOMDIST(24, 27, 0.9, FALSE)

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6
Q

If you were to play the following games 1000 times each, which one of them would be the best choice for maximizing profit over the long term?

Paying $100 for the chance to get back $500 (win $400 + $100 you invested) where your chance of winning is 0.25.

Paying $200 for the chance to get back $1000 (win $800 + $200 you invested) where your chance of winning is 0.2.

Paying $200 for the chance to get back $2000 (win $1800 + $200 you invested) where your chance of winning is 0.1.

Paying $300 for the chance to get back $5000 (win $4700 + $300 you invested) where your chance of winning is 0.05.

A

Paying $100 for the chance to get back $500 (win $400 + $100 you invested) where your chance of winning is 0.25.

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7
Q

The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.5 tickets per day. Interpret the value of the mean.

The number of tickets that is written most often is 6.5 tickets per day.

Half of the days have less than 6.5 tickets written and half of the days have more than 6.5 tickets written.

If we sampled all days, the arithmetic average or expected number of tickets written would be 6.5 tickets per day.

The mean has no interpretation since 0.5 ticket can never be written.

A

If we sampled all days, the arithmetic average or expected number of tickets written would be 6.5 tickets per day.

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8
Q

Which of the following Excel commands would give the probability of 10 or more successes in 25 trials, given that each trial has a 30% chance of success?
=BINOMDIST(10, 25, 0.3, TRUE)
=BINOMDIST(10, 25, 0.3, FALSE)
=BINOMDIST(10, 25, 0.7, TRUE)
=1 - BINOMDIST(10, 25, 0.3, TRUE)
=1 - BINOMDIST(9, 25, 0.3, TRUE)

A

=1 - BINOMDIST(9, 25, 0.3, TRUE)

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9
Q

True or False: Suppose that a judge’s decisions follow a binomial distribution and that his verdict is correct 90% of the time. In his next 10 decisions, the probability that he makes fewer than 2 incorrect verdicts is =BINOMDIST(1,10,0.1,TRUE).

A

True

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10
Q

The following table contains the probability distribution for X = the number of retransmissions necessary to successfully transmit a 1024K data package through a double satellite media.

price: 0 1 2 3
probability: 0.35 0.35 0.25 0.05

Referring to the table above, the probability of at least one retransmission is ________.

A

0.65

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11
Q

A student takes 5 classes and visits exactly 1 professor from 1 of these classes every week for 15 weeks. In other words, the probability that one of these professors will see this student in a given week is 20%. If a particular professor wanted to determine the probability that this student would visit him/her exactly 5 times, then the appropriate calculation would be:
=BINOMDIST(5, 15, 0.2, TRUE)
=BINOMDIST(5, 15, 0.2, FALSE)
=POISSON(5, 3, FALSE)
=NORMSDIST(5, 15, 0.2, FALSE)

A

=BINOMDIST(5, 15, 0.2, FALSE)

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12
Q

The following table contains the probability distribution for X = the number of retransmissions necessary to successfully transmit a 1024K data package through a double satellite media.

X: 0 1 2 3
P(X): 0.35 0.35 0.25 0.05

Referring to the table above, the mean or expected value for the number of retransmissions is ________.

A

1.0

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13
Q

Assume you were given the chance to pay $1000 to play a game. In this game you have a 30% chance to break even and a 40% chance of winning $1000 (You get back $2000). Assuming you can play this game an unlimited number of times, what should you do if you wanted to maximize profit?

Don’t play, the game is a scam!

Play once but be prepared to quit if you lose the first time.

Go get as much money as you can find and play this game every waking hour of your life.

A

Go get as much money as you can find and play this game every waking hour of your life.

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14
Q

In a binomial distribution

the random variable X is continuous.

the probability of event of interest is stable from trial to trial.

the number of trials n must be at least 30.

the results of one trial are dependent on the results of the other trials.

A

the probability of event of interest is stable from trial to trial.

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15
Q

Which of the following about the binomial distribution is not a true statement?

The probability of the event of interest must be constant from trial to trial.

Each outcome is independent of the other.

Each outcome may be classified as either “event of interest” or “not event of interest.”

The random variable of interest is continuous.

A

The random variable of interest is continuous.

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16
Q

If the outcomes of a random variable follow a Poisson distribution, then their

mean equals the standard deviation.

median equals the standard deviation.

mean equals the variance.

median equals the variance.

A

mean equals the variance.

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17
Q

What type of probability distribution will the consulting firm most likely employ to analyze the insurance claims in the following problem?

An insurance company has called a consulting firm to determine if the company has an unusually high number of false insurance claims. It is known that the industry proportion for false claims is 3%. The consulting firm has decided to randomly and independently sample 100 of the company’s insurance claims. They believe the number of these 100 that are false will yield the information the company desires.

Binomial distribution

Poisson distribution

None of these

A

Binomial distribution

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18
Q

What type of probability distribution will most likely be used to analyze warranty repair needs on new cars in the following problem?

The service manager for a new automobile dealership reviewed dealership records of the past 20 sales of new cars to determine the number of warranty repairs he will be called on to perform in the next 90 days. Corporate reports indicate that the probability any one of their new cars needs a warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are independent of one another and is interested in predicting the number of warranty repairs he will be called on to perform in the next 90 days for this batch of 20 new cars sold.

Binomial distribution

Poisson distribution

None of these

A

Binomial distribution

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19
Q

What type of probability distribution will most likely be used to analyze the number of blue chocolate chips per bag in the following problem?

The quality control manager of a candy plant is inspecting a batch of chocolate chip bags. When the production process is in control, the average number of blue chocolate chips per bag is 6.0. The manager is interested in analyzing the probability that any particular bag being inspected has fewer than 5.0 blue chocolate chips.

Binomial distribution

Poisson distribution

None of these.

A

Poisson distribution

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20
Q

The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.5 tickets per day. Interpret the value of the mean.

The number of tickets that is written most often is 6.5 tickets per day.

Half of the days have less than 6.5 tickets written and half of the days have more than 6.5 tickets written.

If we sampled all days, the arithmetic average or expected number of tickets written would be 6.5 tickets per day.

The mean has no interpretation since 0.5 ticket can never be written.

A

If we sampled all days, the arithmetic average or expected number of tickets written would be 6.5 tickets per day.

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20
Q

T or F: The largest value that a Poisson random variable X can have is n.

A

False

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21
Q

For some value of Z, the probability that a standard normal variable is below Z is 0.2090. The value of Z is:

A

-0.81

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22
Q

Given that X is a normally distributed random variable with a mean of 50 and a standard deviation of 2, find the probability that X is between 47 and 54.

A

0.9104

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23
Q

Madame Cleopatra, the telephone psychic, works alone. The duration of her calls is normally distributed with a mean of seven minutes and a standard deviation of two minutes. What Excel calculation would give the fraction of Madame Cleopatra’s calls that last 5 MINUTES OR LESS?

=NORMSINV(-1)
=NORMSINV(5)
=NORMSDIST(5)
=NORMSDIST(-1)

A

=NORMSDIST(-1)

24
Q

The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is ________ that a product is assembled in less than 12 minutes.
=NORMSDIST(-12)
=NORMSDIST(12)
=NORMSDIST(-1.5)
=NORMSDIST(1.5)

A

=NORMSDIST(-1.5)

24
Q

A telephone operator receives calls that have durations which are normally distributed with a mean of 10 minutes and a standard deviation of 3 minutes. 80% of all calls have a duration between 10 – x minutes and 10 + x minutes, where x is the appropriate positive number. What Excel calculation would allow us to find the value of x?
= 2NORMSINV(0.70)
= -3NORMSINV(0.10)
= 2NORMSINV(0.85)
= 3NORMSINV(0.80)

A

= 3*NORMSINV(0.80)

25
Q

The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is ________ that a product is assembled in between 15 and 21 minutes.
=NORMSDIST(21) - 0.5
=NORMSDIST(21) - NORMSDIST(15)
=NORMSDIST(3) - 0.5
=1 - NORMSDIST(3)

A

=NORMSDIST(3) - 0.5

26
Q

In Excel, the command =NORMSDIST(0.4) would return a value of approximately 0.66. What does this mean?

There is about a 66% chance that a standard normal variable takes on a value of 0.4 or less.

There is about a 40% chance that a standard normal variable takes on the value of 0.66 or less.

There is about a 66% chance that a standard normal variable takes on the value of 0.4.

There is about a 66% chance that any normally distributed variable takes on the value of 0.4.

A

There is about a 66% chance that a standard normal variable takes on a value of 0.4 or less.

27
Q

The quantity =NORMSINV(.67) is the same as:
= NORMSINV(.33)
= -1*NORMSINV(.33)
=1-NORMSDIST(.33)
= 1-NORMSINV(.33)
NONE OF THE ABOVE

A

= -1*NORMSINV(.33)

28
Q

If it is given that 90% of all observations for a normally distributed variable have z-scores less than z = 1.29. Then find the probability that z is between 0 and 1.29.

A

0.4

29
Q

The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 110 grams of pyridoxine?

A

0.1554

30
Q

Suppose Z has a standard normal distribution with a mean of 0 and a standard deviation of 1. The probability that Z is more than 0.77 is ________.
0.7794

A

0.2207

31
Q

If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot.

A

0.7745

32
Q

The amount of juice that can be squeezed from a randomly selected orange out of a box of oranges with approximately the same size can most likely be modeled by which of the following distributions?

Poisson distribution

Binomial distribution

Normal distribution

A

Normal distribution

33
Q

T or F: The probability that a standard normal random variable, Z, falls between -2.00 and -0.44 is 0.6472.

A

False

34
Q

T or F: The probability that a standard normal random variable, Z, is less than 5.0 is approximately 0.

A

False

35
Q

T or F: Theoretically, the mean, median, and the mode are all equal for a normal distribution.

A

True

36
Q

T or F: A worker earns $15 per hour at a plant in China and is told that only 2.5% of all workers make a higher wage. If the wage is assumed to be normally distributed and the standard deviation of wage rates is $5 per hour, the average wage for the plant is $7.50 per hour.

A

False

37
Q

T or F: If a data set is approximately normally distributed, its normal probability plot would be S-shaped.

A

false

38
Q

=NORM.S.INV(3.1) gives an error when typed into Excel. This is because

you can’t have a sample with 3.1 observations in it.

all of the observations on the standard normal curve are between -3 and +3.

it makes no sense to talk about a probability of 3.1.

the standard normal distribution is not continuous.

A

it makes no sense to talk about a probability of 3.1.

39
Q

For air travelers, one of the biggest complaints is of the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal and when the flight takes off for these 100 flights.

Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes.

Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes.

Distribution is approximately normal with mean = 10 minutes and standard error = 0.8 minutes.

Distribution is approximately normal with mean = 10 minutes and standard error = 8 minutes.

A

Distribution is approximately normal with mean = 10 minutes and standard error = 0.8 minutes.

40
Q

Sampling distributions describe the distribution of

A

statistics

41
Q

The standard error of the mean:

measures the variability of the mean from sample to sample.

is never larger than the standard deviation of the population.

all of these

decreases as the sample size increases.

A

all of these

42
Q

The mean selling price of new homes in a city over a year was $115,000. The population standard deviation was $25,000. A random sample of 100 new-home sales from this city was then taken. What is the probability that the sample mean selling price was between $113,000 and $117,000?

A

0.58

43
Q

A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, 80% of the samples will have less than ________% of minority students.

A

49

44
Q

The central limit theorem is important because:

the mean of the population can be treated as normally distributed, provided that the sample size used is sufficiently large

all sufficiently large populations can be treated as normally distributed

all sufficiently large samples can be treated as normally distributed

the sampling distribution of the mean can be treated as normally distributed, provided that the sample size used is sufficiently large

A

the sampling distribution of the mean can be treated as normally distributed, provided that the sample size used is sufficiently large

45
Q

If you take a sample from a population and are told that there is a 70% chance that your sample mean will fall between a certain range, then what would you expect to happen if you increased the size (n) of your sample? (Hint: Draw it out!)

There would be less than a 70% chance that the sample mean would fall within this same range

There is not enough information to solve this problem

There would be greater than a 70% chance that the sample mean would fall within this same range

A

There would be greater than a 70% chance that the sample mean would fall within this same range

As you increase the size of the sample the standard deviation (standard error) of the sampling distribution decreases. In other words, as the distribution around the mean decreases a tighter range of values takes up a greater proportion of the data.

46
Q

The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is between 35.95 and 35.98 oz. is ________.

=NORMSDIST(-0.8) - NORMSDIST(-2)
=NORMSDIST(-0.1333)- NORMSDIST(-0.3333)
=NORMSINV(-0.1333) - NORMSINV(-0.3333)
=NORMSINV(-0.8) - NORMSINV(-2)
=1 - NORMSDIST(-0.1333)

A

=NORMSDIST(-0.8) - NORMSDIST(-2)

47
Q

The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. So, 95% of all sample means will be greater than how many grams?

=5NORMSINV(0.95)+110
=5NORMSINV(0.05)+110
=25NORMSINV(0.95)+110
=25NORMSINV(0.05)+110

A

=5*NORMSINV(0.05)+110

48
Q

According to an article, 19% of the entire U.S. population have high-speed access to the Internet. Random samples of size 200 are selected from the U.S. population. Among all these random samples, approximately ________% will have less than 40 people out of 200 who have high-speed access to the Internet.

(HINT: This is a sampling distribution of the proportion problem, but you can easily solve this problem by using simple logic)

A

64

49
Q

The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0. Suppose 36 golfers played the course today. Find the probability that the average score of the 36 golfers exceeded 71.
=NORMSDIST(.333)
=NORMSDIST(2)
=1-NORMSDIST(2)
=1-NORMSDIST(.333)
NONE OF THE ABOVE

A

=1-NORMSDIST(2)

50
Q

T or F: If the amount of gasoline purchased per car at a large service station has a population mean of 15 gallons and a population standard deviation of 4 gallons and a random sample of 4 cars is selected, there is approximately a 68.26% chance that the sample mean will be between 13 and 17 gallons.

A

False

51
Q

T or F: The Central Limit Theorem suggests that the sampling distribution of the sample mean approaches a normal distribution as the sample size increases.

A

True

52
Q

T or F: As the sample size increases, the standard error of the mean increases.

A

False

53
Q

T or F: The standard error of the mean is also known as the standard deviation of the sampling distribution of the sample mean.

A

True

54
Q

T or F: A sampling distribution is defined as the probability distribution of possible sample sizes that can be observed from a given population.

A

False

55
Q

T or F: An unbiased estimator will have a value, on average across samples, equal to the population parameter value.

A

True

56
Q

T or F: As the size of the sample is increased, the standard deviation of the sampling distribution of the sample mean for a normally distributed population will stay the same.

A

False

COB 191 (3 decks)

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