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General Statistics: Ch 4 HW

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created 2 years ago by GreenHero15
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This course is designed to acquaint the student with the principles of descriptive and inferential statistics. Topics will include: types of data, frequency distributions and histograms, measures of central tendency, measures of variation, probability, probability distributions including binomial, normal probability and student's t distributions, standard scores, confidence intervals, hypothesis testing, correlation, and linear regression analysis. This course is open to any student interested in general statistics and it will include applications pertaining to students majoring in athletic training, pre-nursing and business.

updated 2 years ago by GreenHero15

Grade levels:
College: First year, College: Second year, College: Third year, College: Fourth year

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1
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Which of the following values cannot be​ probabilities?

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1.46, 5/3, √2, –0.51

For any event​ A, the probability of A is between 0 and​ 1, inclusive. That​ is, 0 ≤ P(A) ≤ 1.

2

Ten of the 100 digital video recorders​ (DVRs) in an inventory are known to be defective.

What is the probability you randomly select an item that is not​ defective?

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The probability is 0.9.

(100 – 10) ÷ 100

= 90/100 = 0.900

3

Assume that 1100 births are randomly selected and exactly 276 of the births are girls.

Use subjective judgment to determine whether the given outcome is​ unlikely.

Determine whether it is unusual in the sense that the result is far from what is typically expected.

It is unlikely because there are many other possible outcomes that have similar or higher probabilities.

An event is unlikely if its probability is very​ small, such as 0.05 or less. Consider the complement of the outcome and how likely it is.

It is unusual because it is not about 550 as expected.

An event has an unusually low number of outcomes of a particular type or an unusually high number of those outcomes if that number is far from what is typically expected. Consider how many girls would typically be expected in a given random selection of births.

4

For a certain casino slot machine, the odds in favor of a win are given as 29 to 71.

Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.

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The probability is 0.29.​

29 ÷ (29 + 71)

= 29/100 = 0.290

5

You are certain to get a heart, diamond, club, or spade when selecting cards from a shuffled deck.

Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive

The probability is 1.

52 ÷ 52 = 1.000

6
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Refer to the sample data for​ pre-employment drug screening shown below.

If one of the subjects is randomly​ selected, what is the probability that the test result is a false​ positive?

Who would suffer from a false positive​ result? Why?

The probability of a false positive test result is 0.012.

1 ÷ (40 + 1 + 11 + 32)

= 1/84 = 0.0119047619

The person tested would suffer because he or she would be suspected of using drugs when in reality he or she does not use drugs.

7

In a test of a​ gender-selection technique, results consisted of 239 baby girls and 221 baby boys.

Based on this​ result, what is the probability of a girl born to a couple using this​ technique?

Does it appear that the technique is effective in increasing the likelihood that a baby will be a​ girl?

The probability that a girl will be born using this technique is approximately 0.52.

239 ÷ (239 + 221)

= 239/460 = 0.5195652174

No because the technique is effective if the probability of having a girl baby is significantly greater than the expected probability of having a girl​ baby, 0.5

8

In a​ survey, 173 respondents say that they never use a credit​ card, 1221 say that they use it​ sometimes, and 2838 say that they use it frequently.

What is the probability that a randomly selected person uses a credit card​ frequently?

Is it unlikely for someone to use a credit card​ frequently? Consider an event to be unlikely if its probability is less than or equal to 0.05.

How are all of these results affected by the fact that the responses were obtained by those who decided to respond to a survey posted on the​ Internet?

The probability that a randomly selected person uses a credit card frequently is 0.671.

2838 ÷ (173 + 1221 + 2838)

= 2838/4232 = 0.6706049149

No, because the probability of a randomly selected person using a credit card frequently is greater than 0.05.

Since this is a voluntary response​ sample, valid conclusions can only be drawn about the specific group of people who chose to participate.

9

A test for marijuana usage was tried on 150 subjects who did not use marijuana. The test result was wrong 4 times.

a. Based on the available​ results, find the probability of a wrong test result for a person who does not use marijuana.

b. Is it​ "unlikely" for the test to be wrong for those not using​ marijuana? Consider an event to be unlikely if its probability is less than or equal to 0.05.

a. The probability that the test will be wrong is approximately 0.027.

4/150 = 0.0266666667

b. Yes

10

In a survey of consumers aged 12 and​ older, respondents were asked how many cell phones were in use by the household.​ (No two respondents were from the same​ household.) Among the​ respondents, 217 answered​ "none," 288 said​ "one," 371 said​ "two," 152 said​ "three," and 93 responded with four or more. A survey respondent is selected at random.

Find the probability that​ his/her household has four or more cell phones in use.

Is it unlikely for a household to have four or more cell phones in​ use? Consider an event to be unlikely if its probability is less than or equal to 0.05.

​P(four or more cell ​phones) = 0.083

93 ÷ (217 + 288 + 371 + 152 + 93)

= 93/1121 = 0.0829616414

No, because the probability of a respondent with four or more cell phones in use is greater than 0.05.

11
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To the right are the outcomes that are possible when a couple has three children. Refer to that​ list, and find the probability of each event.

a. Among three​ children, there are exactly 3 boys.

b. Among three​ children, there are exactly 0 boys.

c. Among three​ children, there is exactly 1 girl.

d. Among three​ children, there is exactly 2 girls.

a. 1/8

Out of the 8 possible​ outcomes, how many result in exactly 3 boys​?

bbb

b. 1/8

Out of the 8 possible​ outcomes, how many result in exactly 0 boys (3 girls)?

ggg

c. 3/8

Out of the 8 possible​ outcomes, how many result in exactly 1 girl?

gbb, bgb, bbg

d. 3/8

Out of the 8 possible​ outcomes, how many result in exactly 2 girls (1 boy)?

ggb, gbg, bgg

12

If a couple were planning to have three​ children, the sample space summarizing the gender outcomes would​ be: bbb,​ bbg, bgb,​ bgg, gbb,​ gbg, ggb, ggg.

a. Construct a similar sample space for the possible weight outcomes​ (using o for overweight and u for underweight​) of two children.

b. Assuming that the outcomes listed in part​ (a) were equally​ likely, find the probability of getting two overweight children.

c. Find the probability of getting exactly one overweight child and one underweight child.

a. oo, ou, uo, uu

b. 1/4

oo

c. 1/2

ou, uo = 2/4 = 1/2

13

Each of two parents has the genotype brown divided by red​, which consists of the pair of alleles that determine hair color​, and each parent contributes one of those alleles to a child. Assume that if the child has at least one brown ​allele, that color will dominate and the​ child's hair color will be brown.

a. List the different possible outcomes. Assume that these outcomes are equally likely.

b. What is the probability that a child of these parents will have the red divided by red ​genotype?

c. What is the probability that the child will have brown hair color​?

a. b/b, b/r, r/b, r/r

b. 0.25

1/4 = 0.250

c. 0.75

3/4 = 0.750

14

A modified roulette wheel has 44 slots. One slot is​ 0, another is​ 00, and the others are numbered 1 through 42​, respectively. You are placing a bet that the outcome is an odd number.​ (In roulette, 0 and 00 are neither odd nor​ even.)

a. What is your probability of​ winning?

b. What are the actual odds against​ winning?

c. When you bet that the outcome is an odd ​number, the payoff odds are​ 1:1. How much profit do you make if you bet ​$13 and​ win?

d. How much profit should you make on the ​$11 bet if you could somehow convince the casino to change its payoff odds so that they are the same as the actual odds against​ winning?

a. The probability of winning is 21/44.

b. The actual odds against winning are 23​:21.

c. If you win, the payoff is $11.

The payoff odds against event A represent the ratio of net profit​ (if you​ win) to the amount bet.

payoff odds against A = (net ​profit):(amount bet) = 1:1

If the payoff odds are​ a:b and the bet is​ c, the payoff is (a*c)/b . (1*11)/1

d. $12.05

The actual odds against winning are 23​:21​

(23*11)/21 = 12.04761905

15
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In a clinical trial of 2019 subjects treated with a certain​ drug, 24 reported headaches. In a control group of 1579 subjects given a​ placebo, 20 reported headaches.

Denoting the proportion of headaches in the treatment group by pt and denoting the proportion of headaches in the control​ (placebo) group by pc​, the relative risk is pt​/pc. The relative risk is a measure of the strength of the effect of the drug treatment.

Another such measure is the odds​ ratio, which is the ratio of the odds in favor of a headache for the treatment group to the odds in favor of a headache for the control​ (placebo) group, found by evaluating (see image) . The relative risk and odds ratios are commonly used in medicine and epidemiological studies.

Find the relative risk and odds ratio for the headache data.

What do the results suggest about the risk of a headache from the drug​ treatment

The relative risk = 0.938.

relative risk = pt / pc

pt = 24/2019, pc = 20/1579

(24/2019) ÷ (20/1579) = 0.9384843982

The odds ratio = 0938.

[pt / (1 – pt )] ÷ [pc / (1 – pc )]

The drug does not appear to pose a risk of headaches because pt is slightly less than pc.

16

Which of the following is NOT a principle of​ probability?

All events are equally likely in any probability procedure.

It is not always true that all events are equally likely in any probability procedure.

17

Decide whether the following two events are disjoint.

1. Randomly selecting an animal with green eyes

2. Randomly selecting an animal with brown eyes

Yes, because the events cannot occur at the same time.

18

Determine whether the two events are disjoint for a single trial.​ (Hint: Consider​ "disjoint" to be equivalent to​ "separate" or​ "not overlapping.")

Randomly selecting a statistics student and getting someone who brings a notebook to class.

Randomly selecting a statistics student and getting someone who brings a text book to class.

The events are not disjoint. They can occur at the same time.

19

Determine whether the two events are disjoint for a single trial.​ (Hint: Consider​ "disjoint" to be equivalent to​ "separate" or​ "not overlapping.")

Randomly selecting someone who plays soccer.

Randomly selecting someone taking a calculus course.

The events are not disjoint. They can occur at the same time.

20

Find the indicated complement.

A certain group of women has a 0.09​% rate of​ red/green color blindness. If a woman is randomly​ selected, what is the probability that she does not have​ red/green color​ blindness?

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0.9991

(1 – 0.0009)

21
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The following data summarizes results from 1000 ​pre-employment drug screening tests.

If one of the test subjects is randomly​ selected, find the probability that the subject had a positive test result or a negative test result.

P(subject had a positive test result or a negative test ​result) = 1

P(positive test result) + P(negative test result)

= (82 + 83)/1000 + (6 + 829)/1000

= 0.165 + 0.835

=1.00

22
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The following data lists the number of correct and wrong dosage amounts calculated by 32 physicians. In a research​ experiment, a group of 17 physicians was given bottles of epinephrine labeled with a concentration of​ "1 milligram in 1 milliliter​ solution," and another group of 15 physicians was given bottles labeled with a ratio of​ "1 milliliter of a​ 1:1000 solution."

If one of the physicians is randomly​ selected, what is the probability of getting one who calculated the dose​ correctly?

Is that probability as high as it should​ be?

0.5

P(correct dose calculation)

= (12 + 4)/32

= 0.500

No. One would want this probability to be very high.

23
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The following data lists the number of correct and wrong dosage amounts calculated by 32 physicians. In a research​ experiment, a group of 16 physicians was given bottles of epinephrine labeled with a concentration of​ "1 milligram in 1 milliliter​ solution," and another group of 16 physicians was given bottles labeled with a ratio of​ "1 milliliter of a​ 1:1000 solution."

a. For the physicians given the bottles labeled with a​ concentration, find the percentage of correct dosage​ calculations, and then express it as a probability.

b. For the physicians given the bottles labeled with a​ ratio, find the percentage of correct dosage​ calculations, and then express it as a probability.

c. Does it appear that either group did​ better? What does the result suggest about drug​ labels?

a. The probability of a correct dosage calculation given the bottle is labeled with a concentration is 0.750.

P(correct dose calculation with a concentration label)

= 12/16

= 0.75

b. The probability of a correct dosage calculation given the bottle is labeled with a ratio is 0.313.

P(correct dose with a ratio label)

= 5/16

= 0.3125

c. It appears that the group given the labels with concentrations performed better because the probability of a correct dosage calculation for the bottles labeled with a concentration is much greater than the probability of a correct dosage calculation for the bottles labeled with a ratio.This result suggests that labels with concentrations are much better than labels with ratios.

24
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The table below summarizes results from a study of people who refused to answer survey questions. A pharmaceutical company is interested in opinions of the elderly.

What is the probability that the selected subject is someone 60 and over who​ responded?

The probability that the selected subject is someone 60 and over who responded is 0 .161.

P(ages 60 and over who responded)

= (60 & over) ÷ (total responded + total refused)

= 209 ÷ (80 + 262 + 252 + 143 + 145 + 209 + 15 + 24 + 37 + 30 + 39 + 61)

= 0.1611410948

25
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The table below summarizes results from a study of people who refused to answer survey questions. A market researcher is interested in​ responses, especially from those between the ages of 22 and 39.

Find the probability that a selected subject responds or is between the ages of 22 and 39.

The probability that the subject responded or is between the ages of 22 and 39 is 0.896.

---------------------------------------------------

  • P(responded) = (total responded) ÷ (total responded + total refused)

= (75 + 257 + 247 + 138 + 140 + 204) ÷ (75 + 257 + 247 + 138 + 140 + 204 + 11 + 20 + 33 + 26 + 35 + 57) = (1061 / 1243)

  • P(ages 22-39 who r efused )

= (20 + 33) ÷ 1243 = (53 / 1243)

  • P( responded or ages 22-39 ) = (1061 + 53) / 1243

= 0.8962188254

26

Use the following results from a test for marijuana​ use, which is provided by a certain drug testing company. Among 149 subjects with positive test​ results, there are 29 false positive results. Among 155 negative​ results, there are 4 false negative results.

a. How many subjects were included in the​ study?

b. How many subjects did not use​ marijuana?

c. What is the probability that a randomly selected subject did not use​ marijuana?

a. The total number of subjects in the study was 304.

(149 + 155)

b. A total of 180 subjects did not use marijuana

(29 + 155 – 4)

c. The probability that a randomly selected subject did not use marijuana is 0.592.

(180 ÷ 304) = 0.5921052632

27

Use the following results from a test for marijuana​ use, which is provided by a certain drug testing company. Among 148 subjects with positive test​ results, there are 30 false positive​ results; among 154 negative​ results, there are 3 false negative results.

If one of the test subjects is randomly​ selected, find the probability that the subject tested negative or did not use marijuana.​ (Hint: Construct a​ table.)

The probability that a randomly selected subject tested negative or did not use marijuana is 0.609.

P(tested negative or didn't use marijuana)

= (154 / 302) + (30 / 302)

= 184/302 = 0.6092715232

28

Use the following results from a test for marijuana​ use, which is provided by a certain drug testing company. Among 144 subjects with positive test​ results, there are 25 false positive​ results; among 151 negative​ results, there are 3 false negative results.

If one of the test subjects is randomly​ selected, find the probability of a false positive or false negative.​ (Hint: Construct a​ table.)

What does the result suggest about the​ test's accuracy?

The probability of a false positive or false negative is 0.095 .

P(false positive or false negative)

= (25 / 295) + (3/ 295)

= 28/295 = 0.0949152542

With an error rate of 0.095 ​(or 9.5​%), the test does not appear to be highly accurate.

29

Complete the following statement.

P(A or B) indicates​ _______.

the probability that in a single​ trial, event A​ occurs, event B​ occurs, or they both occur.

30

For the given pair of events A and​ B, complete parts​ (a) and​ (b) below.

A: When a page is randomly selected and ripped from a 23​-page document and​ destroyed, it is page 7.​

B: When a different page is randomly selected and ripped from the​ document, it is page 1.

a. Determine whether events A and B are independent or dependent.​ (If two events are technically dependent but can be treated as if they are independent according to the​ 5% guideline, consider them to be​ independent.)

b. Find​ P(A and​ B), the probability that events A and B both occur.

a. The two events are dependent because the occurrence of one affects the probability of the occurrence of the other.

b. The probability that events A and B both occur is 0.0020.

(1/23) x (1/22) = 0.0019762846

31

Consider a bag that contains 223 coins of which 6 are rare Indian pennies. For the given pair of events A and​ B, complete parts​ (a) and​ (b) below.

A: When one of the 223 coins is randomly​ selected, it is one of the 6 Indian pennies.

​B: When another one of the 223 coins is randomly​ selected, it is also one of the 6 Indian pennies.

a. Determine whether events A and B are independent or dependent.

b. Find​ P(A and​ B), the probability that events A and B both occur.

a. The two events are independent because the occurrence of one does not affect the probability of the occurrence of the other.

b. The probability that events A and B both occur is 0.000724.

(6/223) x (6/223) = 0.3

32
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Refer to the table below. Given that 2 of the 220 subjects are randomly​ selected, complete parts​ (a) and​ (b).

a. Assume that the selections are made with replacement. What is the probability that the 2 selected subjects are both group Upper O and type Rh + ​?

a. 0.1356

(81/220) x (81/220) = 0.1355578512

b. 0.1345

(81/220) x (80/219) = 0.1344956413

33

With one method of a procedure called acceptance​ sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. A company has just manufactured 1087 ​CDs, and 467 are defective.

If 3 of these CDs are randomly selected for​ testing, what is the probability that the entire batch will be​ accepted?

Does this outcome suggest that the entire batch consists of good​ CDs? Why or why​ not?

The probability that the whole batch is accepted is 0.185.

(620/1087) x (619/1086) x (618/1085) = 0.185174724

No, because only a probability of 1 would indicate the entire batch consists of good CDs.

34

The principle of redundancy is used when system reliability is improved through redundant or backup components. A​ region's government requires that commercial aircraft used for flying in hazardous conditions must have two independent radios instead of one. Assume that for a typical​ flight, the probability of a radio failure is 0.0052.

What is the probability that a particular flight will be threatened with the failure of both​ radios?

Describe how the second independent radio increases safety in this case.

0.000027

0.0052² = 0.00002704

With one radio there is a 0.0052 probability of a serious​ problem, but with two independent​ radios, the probability of a serious problem decreases substantially. The flight becomes much safer with two independent radios.

35
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The data in the table below summarize results from 112 pedestrian deaths that were caused by accidents.

If two different deaths are randomly selected without​ replacement, find the probability that they both involved intoxicated drivers.

Is such an event​ unlikely?

The probability is 0.561.

(84/112) x (83/111) = 0.560810808

No, because its probability is greater than 0.05.

36

In a market research survey of 2354 ​motorists, 229 said that they made an obscene gesture in the previous month.

a. If 1 of the surveyed motorists is randomly​ selected, what is the probability that this motorist did not make an obscene gesture in the previous​ month?

b. If 50 of the surveyed motorists are randomly selected without​ replacement, what is the probability that none of them made an obscene gesture in the previous​ month? Should the​ 5% guideline be applied in this​ case? Select the correct choice below and fill in the answer box within your choice.

a. The probability is 0.9027.

1 – (229/2354) = 0.9027187766

b. The probability is 0.0060.

(0.9027)50

37

Refer to the figure below in which surge protectors p and q are used to protect an expensive​ high-definition television. If there is a surge in the​ voltage, the surge protector reduces it to a safe level. Assume that each surge protector has a 0.94 probability of working correctly when a voltage surge occurs.

a. If the two surge protectors are arranged in​ series, what is the probability that a voltage surge will not damage the​ television?

b. If the two surge protectors are arranged in​ parallel, what is the probability that a voltage surge will not damage the​ television?

a. 0.9964

1 – (0.06)2

b. 0.8836

0.942

c. The series arrangement provides better protection because it has a higher probability of protection.


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