SHOW ALL NECESSARY WORK. Be Neat and Organized. Good Luck

SHOW ALL NECESSARY WORK. Be Neat and Organized. Good Luck. ... randomly selected, find the probability of getting one ... Find the probability of rand...

14 downloads 619 Views 239KB Size
MTH 157 Sample Test 2 ANSWERS Student_____________________________ Row___ Seat___ M157ST2a Chapters 3 & 4 Dr. Claude S. Moore Score_________ SHOW ALL NECESSARY WORK. Be Neat and Organized. Good Luck. 1.

In a statistics class, 12 students own their own computers and 9 do not. If one of the students is randomly selected, find the probability of getting one who does not own a computer. 9 9 ANSWER: P(not computer) =   0.429 12  9 21

2.

Find each of the following probabilities: a. What is the probability of an event that is certain to NOT occur? ANSWER: P(certain to not occur) = 0.0 b.

What is the probability of an event that is certain to occur? ANSWER: P(certain to occur) = 1.0

c.

A sample space consists of 14 separate events that are equally likely to occur. What is the probability of each? 1 ANSWER: P(each) = = 0.0714 14

d.

On a true/false test, what is the probability of answering a question correctly if you make a random guess? ANSWER: P(correct guess) = 0.5

e.

On a multiple-choice test with five possible answers for each question, what is the probability of answering a question correctly if you make a random guess? 1 ANSWER: P(correct guess) = = 0.2 5

3.

Baseball player Mark McGwire broke a record when he hit 70 home runs in the 1998 season. During that season, he was at bat 509 times. If one of those “at bats” is randomly selected, find the probability that it is one of the times he did not hit a home run. 70 439  ANSWER: P(not home run) = 1  = 0.862 509 509

4.

A roulette wheel has 38 slots: One slot is 0, another slot is 00, and the other slots are numbered 1 through 36. a. If you bet all of your textbook money on the number 15, what is the probability that you will win? 1 ANSWER: P(win) = = 0.026 38 b. Is it “unusual” to win when you bet on a single number in roulette? Why? ANSWER: Yes, because 0.026 < 0.05

5.

In a Bruskin-Goldring Research poll, respondents were asked how a fruitcake should be used. One hundred forty-five respondents indicated that it should be used for a doorstop, and 800 other respondents cited other uses, including birdfeed, landfill, and a gift. If one of these respondents is randomly selected, what is the probability of getting someone who would use the fruitcake as a doorstop? ANSWER: P(doorstop) =

6.

145 145   0.153 145  800 945

Find each of the following probabilities: a. If P( A ) = 0.731, find P(A). ANSWER: P(A) = 1 – 0.731 = 0.269 b.

Based on data from the National Health Examination, the probability of a randomly selected adult male being 6 ft or shorter is 0.86. Find the probability of randomly selecting an adult male and getting someone taller than 6 ft. ANSWER: P(>6 ft) = 1 – 0.86 = 0.14

7.

In a study of 82 young (under the age of 32) drivers, 39 were men who were ticketed, 11 were men who were not ticketed, 8 were women who were ticketed, and 24 were women who were not ticketed (based on data from the Department of Transportation). If one of these subjects is randomly selected, find the probability of getting a man or someone who was ticketed. ANSWER: P(man or ticketed) =

8.

50  8 58   0.707 82 82

Determine whether the following events are mutually exclusive: a. If P(A) = 4/15, P(B) = 7/15, and P(A or B) = 10/15, what do you know about events A and B? 10 4 7 11    ANSWER: NO, because P(A or B) = 15 15 15 15 b. If P(A) = 2/ 7, P(B) = 4/ 7, and P(A or B) = 6/ 7, what do you know about events A and B? 6 2 4   7 7 7 A pool of potential jurors consists of 10 men and 12 women. If two different people are randomly selected from this pool, find the probability that they are both women.

ANSWER: YES, because P(A or B) = 9.

ANSWER: P(WW) =

12 11 4 2     0.286 22 21 14 7

10. A quick quiz consists of five multiple-choice questions, each with four possible answers, only one of which is correct. If you make random guesses for each answer, what is the probability that all five of your answers are wrong? 3 3 3 3 3 243 ANSWER: P(wrong) =       0.237 4 4 4 4 4 1024 11. A classic excuse for a missed test is offered by four students who claim that their car had a flat tire. On the makeup test, the instructor asks the students to identify the particular tire that went flat. If they really didn’t have a flat tire and randomly select one that supposedly went flat, what if the probability that they will all select the same tire? 4 1 1 1 1 ANSWER: P(same) =      0.016 4 4 4 4 64 12. Does P(x) = x/6 (where x can take on the values of 0, 1, 2, 3) describe a probability distribution? Explain your answer. ANSWER: X P(X) 0 0 1 1/6 2 2/6 3 3/6 Total 1.00 Yes, 0  P(X)  1 and sum of P(X’s) = 1. 13. To settle a paternity suit, two different people are given blood tests. If x is the number having group A blood, then x can be 0, 1, or 2, and the corresponding probabilities are 0.36, 0.48, and 0.16, respectively (based on data from the Greater New York Blood Program). Determine whether the above information describes a probability distribution. If it does not, explain why. If it does, find its mean. ANSWER: X P(X) XP(X) 0 0.36 0.00 1 0.48 0.48 2 0.16 0.32 Total 1.00 0.80 Yes, 0  P(X)  1 and sum of P(X’s) = 1. Mean = 0.80. 14. When you give a casino $10 for a bet on the “pass line” in the game of craps, there is a 244/495 probability that you will win $10 and a 251/495 probability that you will lose $10. What is your expected value?  244   251  10 244  251  0.1414 ANSWER: Expected value = E(win) = 10   10   495   495  495

15. Given that n = 6 and p = 0.95 (in a binomial probability distribution), find the following probabilities: a. The probability of exactly 2 successes. ANSWER: P(X=2) = 0.000085 b. The probability of at least 2 successes. ANSWER: P(X≥2) = 0.99998 16. The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, Southwest Air had the best rate with 80% of its flights arriving on time. A test is conducted by randomly selecting 15 Southwest flights and observing whether they arrive on time. a. Find the probability that exactly 8 flights arrive on time. ANSWER: P(X=8) = 0.0138 b. Would it be unusual for Southwest to have 6 flights arrive late? Why or why not? ANSWER: P(6 late) = 0.043 YES, because 0.043 < 0.05 17. Several students are unprepared for a multiple-choice test with 20 questions, and all of their answers are guesses. Each question has four possible answers, and only one of them is correct. a. Find the mean for the number of correct answers for such students. ANSWER:  = np = 20(1/4) = 5 b.

Find the standard deviation for the number of correct answers for such students.

ANSWER:  =

 1  3  20    1.936  4  4 

c.

Would it be unusual for a student to pass by guessing and getting at least 12 correct answers? Why or why not? ANSWER: P(X≥12) = 1 – P(X≤11) = 1 – 0.99906 = 0.00094 YES, because 0.00094 < 0.05 18. Mars, Inc., claims that 10% of its M&M plain candies are blue, and a sample of 150 such candies is randomly selected. a. Find the mean for the number of blue candies in such groups of 150. ANSWER:  = 150(0.1) = 15 b. Find the standard deviation for the number of blue candies in such groups of 150. ANSWER:  = 150(0.1)(0.9)  3.674

For Problems 1 and 2, assume that voltages in a circuit vary between 5 volts and 11 volts, and voltages are spread evenly over the range of possibilities (Uniform Distribution). Find the probability of the given range of voltage levels. [Make a sketch for each problem.] 19.

Greater than 9 volts

P(X)

20.

Between 5.5 volts and 10 volts

P(X)

. . . . . . . . . . . . . Number of Volts

P(X≥9) = P(9≤X≤11) =

11  9 2 1   11  5 6 3

. . . . . . . . . . . . . Number of Volts

P(5.5≤X≤10) =

10  5.5 4.5 45 3    11  5 6 60 4

For Problems 21, 22, 23, and 24, assume that the readings on the thermometers are normally distributed with a mean of 0C and a standard deviation of 1.00C. [Make a sketch for each problem.] 21.

Find the indicated probability where z is the reading in degrees: P(z < 1.645) This is a standard normal distribution. Thus, use the normal distribution table to find the area to the left of z = 1.645. P(z<1.645) =

1.645

22.

Find the indicated probability where z is the reading in degrees: P(1.96 < z < 2.33) Since this is a standard normal distribution, use the normal distribution table to find the area between z = 1.96 and z = 2.33. P(1.96
23.

Find the temperature reading corresponding to P20, the 20th percentile.

24. If 12% of the thermometers are rejected because they have readings that are too high, but all other thermometers are acceptable, find the reading that separates the rejected thermometers from the others. [Make a sketch for each problem.]

2.33

For Problems 25, 26, 27, and 28, assume that women’s weights are normally distributed with a mean given by  = 143 lbs and standard deviation given by  = 29 lbs. Also, assume that a woman is randomly selected. Let x = weight in pounds. 25. Find the indicated probability: P(150 < x < 180)

26. Find the indicated probability: P(x < 186.5)

27. Find the sixth decile, D6, which is the weight separating the bottom 60% from the top 40%.

28.a)

If 1 woman is randomly selected, find the probability that her weight is above 140 lb.

b) If 100 women are randomly selected, find the probability that they have a mean weight greater than 140 lb.

29.The U.S. Army requires women’s heights to be between 58 in. and 80 in. Find the percentage of women meeting that height requirement. Are many women being denied the opportunity to join the Army because they are too short or too tall? [Assume that heights of women are normally distributed with a mean of  = 63.6 in. and a standard deviation of  = 2.5 in.]

30.The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. If we stipulate that a baby is premature if born at least three weeks early, what percentage of babies is born prematurely? Three weeks early would be 3(7) = 21 days early. This would be on or before day 247 (268 – 21). About 8.1% of babies would be born prematurely.

For Problems 31 and 32, use the fact that IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. [Make a sketch for each problem.] 31.a)

People are considered to be “intellectually very superior” if their score is above 130. What percentage of people fall into that category? About 2.3% of people would be considered “intellectually very superior”.

b) If we redefine the category of “intellectually very superior” to be scores in the top 2%, what does the minimum score become? The top 2% means 98% would be below the IQ score of 130.8.

32.If 25 people are randomly selected for an IQ test, find the probability that their mean IQ score is between 95 and 105. The probability that the mean IQ score of 25 people is 0.9044.

Comments: I hope this sample test and these answers are helpful in preparing for your test.