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Math 251, Final Review Questions
Name:
ANSWERS (to questions 5,6,9 and 15).
5. Consider the data (which are systolic blood pressures of 25 subjects):
105
126
146
108
126
152
110
128
166
110
130
188
112
130
190
112
130
116
132
118
134
118
136
120
140
(a) What class width should be chosen if you would like to have 8 classes?
ANS: First, the range divided by the number of classes is (190 – 105)/8 = 10.625. Now go up to the
next who number to make sure all data are covered in the 8 classes, so we choose a class width of 11.
(b) Complete the following table for this data given that the first class has limits 105—119
Lower
Limit
Upper
Lower
Upper
Cumulative Relative
Limit Boundary Boundary Midpoint Frequency Frequency Frequency
105
119
104.5
119.5
112
9
9
.36
120
134
119.5
134.5
127
9
18
.36
135
149
134.5
149.5
142
3
21
.12
150
164
149.5
164.5
157
1
22
.04
165
179
164.5
179.5
172
1
23
.04
180
194
179.5
194.5
187
2
25
.08
(c) Find the median, Q1, and Q3 for the above data. Draw a box and whisker plot for the data. You may
draw it horizontally if you prefer.
ANS: The median is in the (25 + 1)/2th place, i.e. the 13th place, so the median is 128.
The first quartile is the median of the first 12 numbers, which is 114 (average of 6th and 7th data).
The third quartile is the median of the highest 12 numbers which is 138 (average of 19th and 20th data).
High: 190
Third Quartile: 138
Median: 128
First Quartile: 114
Low: 105
See text for method of constructing box and whisker plot. The lower whisker starts at 105 and goes to
114, the low edge of the box is at 114, the upper edge is at 138, and the line in the box is at 128. The
upper whisker starts at 138 and goes up to 190.
(d) Construct a relative frequency histogram for the data using the table in (b).
Relative Frequency Histogram
0.4
0.35
Relative Frequency
0.3
0.25
0.2
0.15
0.1
0.05
0
5
4.
19
5
9.
17
5
4.
16
5
9.
14
5
4.
13
5
9.
11
5
4.
10
Systolic Pressure
6. Consider the data (which are systolic blood pressures of 50 subjects):
100,102,104,108,108,110,110,112,112,112,115,116,116,118,118,
118,118,120,120,126,126,126,128,128,128,130,130,130,130,130,
132,132,134,134,136,136,138,140,140,146,148,152,152,152,156,
160,190,200,208,208
(a) What class width should be chosen if you would like to have 6 classes.
ANS: (208 – 100)/6 = 18. Go to the next higher whole number to ensure that all of the data is
covered. Thus a class width of 19 would be suitable.
(b) Suppose you don’t want a class width of 19, but would like a class width of 15 irrespective of how
many classes that would give you. Complete the following table for this data.
Lower
Limit
Upper
Limit
Lower
Boundary
Upper
Boundary
Midpoint
Frequency
Cumulative Relative
Frequency Frequency
100
114
99.5
114.5
107
10
10
.20
115
129
114.5
129.5
122
15
25
.30
130
144
129.5
144.5
137
14
39
.28
145
159
144.5
159.5
152
6
45
.12
160
174
159.5
174.5
167
1
46
.02
175
189
174.5
189.5
182
0
46
.00
190
204
189.5
204.5
197
2
48
.04
205
219
204.5
219.5
212
2
50
.04
(c) Draw a frequency histogram using the table in (b).
Histogram
16
14
Frequenc y
12
10
8
Fre quen cy
6
4
2
0
4.
14
9.
15
4.
17
9.
18
4.
20
9.
21
5
5
5
5
5
5
5
5
9.
12
4.
11
.5
99
Systolic Blood Pres sure
(d) Draw a frequency polygon using the table in (b).
Frequenc y Polygon for B.P.
16
14
Frequency
12
10
8
6
4
2
0
92
107
122
137
152
167
182
197
212
227
Sy stolic Press ure
(e) Draw an Ogive using the table in (b).
Ogiv e
60
Cumulative Frequency
50
40
30
20
10
0
99.5
114.5
129.5
144.5
159.5
174.5
Sy stolic Press ure
189.5
204.5
219.5
f) Find the median, mode, range and first and third quartiles for the data in this problem.
Range: 108
Third Quartile: 140 (the median of the largest 25 numbers is in the 38th place)
First Quartile: 116 (the median of the first 25 numbers in the 13th place)
Median: 129 (the median is the average of the 25th and 26th places)
Mode: 130 (the most commonly occurring data)
9. (a) Make a stem and leaf display for the following data.
58
92
85
52
66
84
68
68
90
86
87
57
72
86
77
66
73
76
97
61
84
89
70
93
84
75
58
91
72
47
91
73
ANS: (not exactly a stem and leaf display, but a similar table)
4
5
6
7
8
9
7
2788
16688
02233567
44456679
011237
4|7 = 47
(b) After making the display, find the median of the data.
ANS: The median is the average of the 16th and 17th places which is 75.5
15. (a) (1 pt) Fill in the missing probability for the following discrete random variable:
X
3
6
9
10
12
P(x)
.10
.15
.25
?
.11
ANS:
? = 1 - 0.61 = 0.39
(b) The number of cars per household in a small town is given by
Cars
Households
0
20
1
280
2
75
3
25
(i) Make a probability distribution for x where x represents the number of cars per household in this
small town.
ANS:
Cars (x)
0
1
2
3
P(x)
.0500
.7000
.1875
.0625
(ii) Find the mean and standard deviation for the random variable in (i)
ANS:
Mean:
 = 0(.0500)+1(.7000)+2(.1875)+3(.0625) = 1.2625
Variance:
2 = 1(.7000)+4(.1875)+9(.0625) – 1.26252 = 0.41859375
Standard Deviation:  = (0.41859375)1/2 = 0.64699
(iii) What is the average number of cars per household in that small town? Explain what you mean by
average.
ANS:
On average, there are 1.2625 cars per household; this is the mean number of cars per household.
This is the number one would get if they took the total number of cars in the town and divided
by the total number of households.
(c) Different Random Variable Question: (From p. 219 #15) Combinations of Random Variables. Norb
and Gary entered in a local golf tournament. Both have played the local course many times. Their
scores are random variables with the following means and standard deviations.
Norb, x1: 1 = 115;
1 = 12
Gary, x2: 2 = 100;
2 = 8
Assume that Norb’s and Gary’s scores vary independently of each other. The difference between their
scores is W = x1- x2. Compute the mean, variance and standard deviation for the random variable W.
ANS:
Mean:  = 115 – 100 = 15
Variance: 2 = 12122 + 1282 = 144 + 64 = 208.
Standard Deviation:   14.42.
(d) Identify the following random variables as continuous or discrete.
(i)
(ii)
x = the winning time of a horse race
x = the numbers of customers in Vons at a given day
(iii)
(iv)
(v)
x = the number of traffic accidents in LA County in a given year
x = weight of the winning Jockey at a horse race
x = the length of time it takes a person to drive from Anaheim to Hesperia.
ANS: (i),(iv) and (v) are continuous; (ii) and (iii) are discrete.