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Elementary Statistics: Exam 2: Lessons 3 & 4

Module by: Susan Dean, Dr. Barbara Illowsky

Summary: Practice final exam for use with Collaborative Statistics (col10522) by Barbara Illowsky and Susan Dean.

Questions 1 – 4 refer to the following:

The following table classified eighty-two children by age and favorite meals at a McDonald's ® restaurant.

Age (in years) against favorite foods
  6 or below 7-12 13-15 Total
Hamburger 6 12 16 34
Chicken Nuggets 9 13 11 33
Filet ‘o’ fish 2 5 8 15
Totals 17 30 35 82

Assume a child is randomly selected.

Exercise 1

Find the probability of being “7-12” years old AND preferring “chicken nuggets”

  • A. 13821382 size 12{ { {"13"} over {"82"} } } {}
  • B. 28822882 size 12{ { {"28"} over {"82"} } } {}
  • C. 33823382 size 12{ { {"33"} over {"82"} } } {}
  • D. 13611361 size 12{ { {"13"} over {"61"} } } {}

Solution 1

A

Exercise 2

Find the probability of being “13-15” years old OR preferring “Filet’o’fish”.

  • A. 882882 size 12{ { {8} over {"82"} } } {}
  • B. 50825082 size 12{ { {"50"} over {"82"} } } {}
  • C. 851851 size 12{ { {8} over {"51"} } } {}
  • D. 42824282 size 12{ { {"42"} over {"82"} } } {}

Solution 2

D

Exercise 3

Find the probability of “preferring Hamburger” given that the randomly selected child is 13-15 years old.

  • A. 16821682 size 12{ { {"16"} over {"82"} } } {}
  • B. 16351635 size 12{ { {"16"} over {"35"} } } {}
  • C. 16341634 size 12{ { {"16"} over {"34"} } } {}
  • D. 16701670 size 12{ { {"16"} over {"70"} } } {}

Solution 3

B

Exercise 4

The events “preferring Hamburger” and “being 13-15 years old” are:

  • A. Mutually exclusive
  • B. Independent
  • C. Neither mutually exclusive or independent.
  • D. Both mutually exclusive and independent.

Solution 4

C

Exercise 5

EE size 12{E} {} and FF size 12{F} {} are two events such that P(E)=0.60P(E)=0.60 size 12{P \( E \) =0 "." "60"} {}, P(EP(E size 12{P \( E} {} or F)=0.90F)=0.90 size 12{F \) =0 "." "90"} {} and P(EP(E size 12{P \( E} {} and F)=0.50F)=0.50 size 12{F \) =0 "." "50"} {}. Find P(F)P(F) size 12{P \( F \) } {}.

  • A. 0.80
  • B. 0.30
  • C. 0.40
  • D. 0.10

Solution 5

A

Exercise 6

The probability that a randomly chosen adult resident of Bayview city owns a boat is 0.16. The probability that a randomly chosen adult rents an apartment is 0.30. The probability that the adult owns a boat given he/she rents an apartment is 0.20.

  • A. 0.048
  • B. 0.24
  • C. 0.10
  • D. 0.06

Solution 6

D

Exercise 7

Possessing a boat and renting an apartment are:

  • A. independent events
  • B. mutually exclusive
  • C. both independent and mutually exclusive
  • D. neither independent nor mutually exclusive

Solution 7

D

Questions 8 – 9 refer to the following:

A bag contains 4 red marbles and 5 blue marbles. Two marbles are randomly drawn without replacement.

Exercise 8

Find the probability of the event “The first marble is red and the second is blue.”

  • A. 20812081 size 12{ { {"20"} over {"81"} } } {}
  • B. 20722072 size 12{ { {"20"} over {"72"} } } {}
  • C. 412412 size 12{ { {4} over {"12"} } } {}
  • D. 4949 size 12{ { {4} over {9} } } {}

Solution 8

B

Exercise 9

Find the probability that both marbles are red.

  • A. 16811681 size 12{ { {"16"} over {"81"} } } {}
  • B. 781781 size 12{ { {7} over {"81"} } } {}
  • C. 12721272 size 12{ { {"12"} over {"72"} } } {}
  • D. 872872 size 12{ { {8} over {"72"} } } {}

Solution 9

C

Exercise 10

Approximately 70% of U. S. adults had at least one pet as a child. We randomly survey 60 U. S. adults. We are interested in the number that had at least one pet as a child. The probability that at least 3 adults had at least one pet as a child means:

  • A. P(X=0)+P(X=1)+P(X=2)P(X=0)+P(X=1)+P(X=2) size 12{P \( X=0 \) +P \( X=1 \) +P \( X=2 \) } {}
  • B. P(X=0)+P(X=1)+P(X=2)+P(X=3)P(X=0)+P(X=1)+P(X=2)+P(X=3) size 12{P \( X=0 \) +P \( X=1 \) +P \( X=2 \) +P \( X=3 \) } {}
  • C. P(X=4)+P(X=5)+P(X=6)+...+P(X=60)P(X=4)+P(X=5)+P(X=6)+...+P(X=60) size 12{P \( X=4 \) +P \( X=5 \) +P \( X=6 \) + "." "." "." +P \( X="60" \) } {}
  • D. P(X=3)+P(X=4)+P(X=5)+...+P(X=60)P(X=3)+P(X=4)+P(X=5)+...+P(X=60) size 12{P \( X=3 \) +P \( X=4 \) +P \( X=5 \) + "." "." "." +P \( X="60" \) } {}

Solution 10

D

Questions 11 – 12 refer to the following:

A plumber has determined the possible number of house calls to be made each day, and their related probabilities:

x=x= size 12{x={}} {}# of house calls P ( x ) P ( x ) size 12{P \( x \) } {}
0 0.10
1 0.40
2 0.25
3 0.15
4 0.10

Exercise 11

What is the probability that he makes at least 1, but no more than 3 house calls in a day?

  • A. 0.65
  • B. 0.80
  • C. 0.50
  • D. 0.40

Solution 11

B

Exercise 12

If the plumber charges a flat fee of $ 40 for a house call, the expected daily income is:

  • A. $70
  • B. $175
  • C. $400
  • D. $1.75

Solution 12

A

Exercise 13

If X~B(40,0.2)X~B(40,0.2) size 12{X "~" B \( "40",0 "." 2 \) } {}, then P(X>11)=P(X>11)= size 12{P \( X>"11" \) ={}} {}

  • A. 0.0432
  • B. 0.0001
  • C. 0.0875
  • D. 0.1608

Solution 13

C

Exercise 14

The Fizz–Full Soda Company knows that 4% of the bottles of soda it produces are filled with less soda than required. If one purchases 10 bottles at random, the probability that at most 2 of these bottles will have less soda than required is:

  • A. 0.0519
  • B. 0.9938
  • C. 0.9418
  • D. 0.0080

Solution 14

B

Questions 15 – 16 refer to the following:

Assume the statistics final is a multiple-choice exam with 50 questions, each question having 5 choices, only one of which is correct. Assume you answer all questions at random (guessing).

Exercise 15

The expected number of questions you would get correct would be:

  • A. 5
  • B. 10
  • C. 40
  • D. 45

Solution 15

B

Exercise 16

Based upon numerical calculations, would you be surprised if a person got exactly half of the questions correct?

  • A. yes, because it is impossible
  • B. yes, because the probability is almost 0
  • C. no, because the probability is 0.50
  • D. no, because it is the most likely probability

Solution 16

B

Exercise 17

If sampling without replacement occurs, do the picks follow the Binomial Distribution?

  • A. Yes, because each pick is independent from the others.
  • B. No, because the probability of success on each pick changes.
  • C. Yes, if we are counting the number of successes.
  • D. No, because we may not have any successes.

Solution 17

B

Exercise 18

Ninety-four percent of California community college transfers feel that their community college adequately prepared them to handle upper-division coursework at their transfer university. We randomly survey 14 California community college transfers. We are interested in the number that feel that their community college adequately prepared them to handle upper division coursework at their transfer university. List the values that X X size 12{X} {} , the Random Variable, may take on.

  • A. 1,2,3,...,141,2,3,...,14 size 12{1,2,3, "." "." "." ,"14"} {}
  • B. 1,2,3,...,941,2,3,...,94 size 12{1,2,3, "." "." "." ,"94"} {}
  • C. 0,1,2,...,140,1,2,...,14 size 12{0,1,2, "." "." "." ,"14"} {}
  • D. 0,1,2,...,940,1,2,...,94 size 12{0,1,2, "." "." "." ,"94"} {}

Solution 18

C

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