0.2 Practice tests (1-4) and final exams  (Page 5/36)

37 . You had an unusually long wait time, which is bad: 82 percent of patients had a shorter wait time than you, and only 18 percent had a longer wait time.

2.4: box plots

38 . 5

39 . 3

40 . 7

41 . The median is 86, as represented by the vertical line in the box.

42 . The first quartile is 80, and the third quartile is 92, as represented by the left and right boundaries of the box.

43 . IQR = 92 – 80 = 12

44 . Range = 100 – 75 = 25

2.5: measures of the center of the data

45 . Half the runners who finished the marathon ran a time faster than 3:35:04, and half ran a time slower than 3:35:04. Your time is faster than the median time, so you did better than more than half of the runners in this race.

46 . 61.5, or $61,500

47 . 49.25 or $49,250

48 . The median, because the mean is distorted by the high value of one house.

2.6: skewness and the mean, median, and mode

49 . c

50 . a

51 . They will all be fairly close to each other.

2.7: measures of the spread of the data

52 . Mean: 15
Standard deviation: 4.3
$\mu =\frac{10+11+15+15+17+22}{6}=15$
$s=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}=\sqrt{\frac{94}{5}}=4.3$

53 . 15 + (2)(4.3) = 23.6

54 . 13.7 is one standard deviation below the mean of this data, because 15 – 4.3 = 10.7

55 . $z=\frac{95-85}{5}=2.0$
Susan’s z -score was 2.0, meaning she scored two standard deviations above the class mean for the final exam.

3.1: terminology

56 . $P(B)=\frac{25}{90}=0.28$

57 . Drawing a red marble is more likely.
$P(R)=\frac{50}{80}=0.62$
$P(Y)=\frac{15}{80}=0.19$

58 . P ( F AND S )

59 . P ( E | M )

3.2: independent and mutually exclusive events

60 . P ( A AND B ) = (0.3)(0.5) = 0.15

61 . P ( C OR D ) = 0.18 + 0.03 = 0.21

3.3: two basic rules of probability

62 . No, they cannot be mutually exclusive, because they add up to more than 300. Therefore, some students must fit into two or more categories (e.g., both going to college and working full time).

63 . P ( A and B ) = ( P ( B | A ))( P ( A )) = (0.85)(0.70) = 0.595

64 . No. If they were independent, P ( B ) would be the same as P ( B | A ). We know this is not the case, because P ( B ) = 0.70 and P ( B | A ) = 0.85.

3.4: contingency tables

65 .

66 . $P\text{(honor roll|study at least 15 hours word per week) = }\frac{482}{1000}=0.482$

67 . $P(\text{studies less than 15 hours word per week)}=\frac{125+193}{1000}=0.318$

68 . Let P ( S ) = study at least 15 hours per week
Let P ( H ) = makes the honor roll
From the table, P ( S ) = 0.682, P ( H ) = 0.607, and P ( S AND H ) =0.482.
If P ( S ) and P ( H ) were independent, then P ( S AND H ) would equal ( P ( S ))( P ( H )).
However, ( P ( S ))( P ( H )) = (0.682)(0.607) = 0.414, while P ( S AND H ) = 0.482.
Therefore, P ( S ) and P ( H ) are not independent.

3.5: tree and venn diagrams

69 .

70 .

Practice test 2

4.1: probability distribution function (pdf) for a discrete random variable

Use the following information to answer the next five exercises. You conduct a survey among a random sample of students at a particular university. The data collected includes their major, the number of classes they took the previous semester, and amount of money they spent on books purchased for classes in the previous semester.

1. If X = student’s major, then what is the domain of X ?