# Math

```1. Five cards are drawn from an ordinary deck without replacement. Find
the probability of getting
a. All red cards
π(πππ πππ πππππ ) =
26 25 24 23 22
7,893,600
253
&times;
&times;
&times;
&times;
=
ππ
52 51 50 49 48
311,875,200
9996
b. All diamonds
π(πππ ππππππππ ) =
13 12 11 10 9
154,440
33
&times;
&times;
&times;
&times;
=
ππ
52 51 50 49 48
311,875,200
66640
c. All aces
π(πππ πππ) =
4
3
2
1
24
1
&times;
&times;
&times;
=
ππ
52 51 50 49
311,875,200
12994800
2. For the given data below.
102
168
168
116
179
186
113
170
117
132
160
108
128
163
171
117
187
173
156
185
161
182
158
163
183
163
168
171
167
182
a. Construct a grouped frequency distribution. Use six classes.
Range = Highest Value - Lowest Value
= 187 - 102
= 85
Class
Limits
102 - 116
116 - 129
130 - 143
144 - 157
158 - 171
172 - 187
Freque
ncy
4
3
1
1
13
8
Cumulative
Frequency
4
7
8
9
22
30
b. Draw a histogram and frequency polygon for the frequency distribution
obtained from the data.
Class
Limits
102 - 116
116 - 129
130 - 143
144 - 157
158 - 171
172 - 187
Class
Midpoint
109
123
137
144
165
179
Frequen
cy
4
3
1
1
13
8
FREQUENCY
14
12
10
8
6
4
2
0
13
8
4
ΠΠΠΠΠΠΠΠ ΠΠ‘Π
109
3
123
1
1
137 144 165
CLASS MIDPOINT
179
14
12
10
8
6
4
2
0
109
123
137
144
165
179
ΠΠΠΠΠΠΠΠ ΠΠ‘Π
c. Find the mean, median, mode, mid-range, range, variance and standard
deviation for the data.
102 116 128 158 163 167 168 171 182 185
108 117 132 160 163 168 170 173 182 186
113 117 156 161 163 168 171 179 183 187
ππ’π ππ π‘ππππ
mean =
π
4967
mean = 30
mean = 156.5666667
Median = 165
Mode = 168
Standard Deviation, σ: 25.85
Count, N:
30
Sum, Σx:
4697
Mean, μ:
156.56666666667
Variance, σ2:
668.44555555556
π΄(π₯π − π)2
σ2 =
π
σ2 =
(102 − 156.56666666667)2 + ...+ (182 − 156.56666666667)2
20053.366666667
σ2 =
30
σ2 = 668.44555555556
σ=
√668.44555555556
SD = 25.854314060821
30
3. On the daily run of an express bus, the average number of passengers is
48. The standard deviation is 3. Assume the data are normally distributed.
Find the probability that the bus will have
a. Between 36 and 40 passengers
b. Fewer than 42 passengers
c. More than 48 passengers
d. Between 43 and 47 passengers.
4. The data represent the heights in feet and the number of stories of the
tallest buildings in Cleveland.
a. Draw a scatter plot.
y
60
No. of stories
50
40
30
y
20
10
0
0
200
400
600
800
1000
height of Building
b. Find the value of r.
c.
x
y
947
708
658
529
450
446
430
420
xy
57
52
46
40
31
28
24
26
53979
36816
30268
21160
13950
12488
10320
10920
x2
y2
896809
3249
501264
2704
432964
2116
279841
1600
202500
961
198916
784
184900
576
176400
676
419
5007
9
∑
n
π=
π=
π=
π=
π=
π=
32
13408
175561
1024
336 203309 3049155 13690
n(∑xy) − (∑x)(∑y)
√[n∑x 2 − (∑x) 2 ][n∑y 2 − (∑y) 2 )]
9(203,309) − (5,007)(336)
√9(3,049,155) − (5007) 2 ][9(13,690) − (336) 2 )]
1,829,781 − 1,682,352
√27,442,395 − 25,070,049][123,210 − 112,896]
147,429
√(2,372,346)(10,314 )
147,429
√24,468,376,644
147,429
156,424
π = 0.942497792 or 0.94
d. Test the significance of r at the 5% level and at the 1% level.
d. Find the equation of the regression line and draw the line on the scatter
plot, but only if r is significant.
y
70
y = 0,0621x + 2,7601
R&sup2; = 0,8883
No. of stories
60
50
40
30
y
20
ΠΠΈΠ½Π΅ΠΉΠ½Π°Ρ (y)
10
0
0
200
400
600
800
height of Building
1000
e. Describe the nature of the relationship if one exists.
There is sufficient evidence to conclude that there is a significant linear
relationship between x and y because the correlation coefficient is
significantly different from zero.
f. Predict the number of stories in a 500 – feet building.
Y = 0.0621x + 2.7601
Y = 0.0621(500)+2.7601
Y = 31.05+2.7601
Y = 33.8101
```