Download + 32.5(10)

REVIEW OF UNIT 1
Number of Videos
Rented
3
Number of Families
4
6
8
10
16
14
11
2
4
1) The table displays the number of videos rented.
a. How many families were surveyed?
16 + 14 + 11 + 2 + 4 = 47
REVIEW OF UNIT 1
Number of Videos
Rented
3
4
6
8
10
Number of Families
16
14
11
2
4
1) The table displays the number of videos rented.
b. Find the median number of videos rented by
these families.
Median # of families: 47+1 / 2= 48 / 2 = 24
Since 16 families rented 3 videos, the 24 families
would fall into the 4- video category. 16 + 14 = 30
REVIEW OF UNIT 1
Number of Videos
Rented
3
4
6
8
10
Number of Families
16
14
11
2
4
1) The table displays the number of videos rented.
c. Find the mean number of videos rented by
these families.
3(16)  4(14)  6(11)  8( 2)  10(4) 226

 4.81
47
47
2) The histogram below shows the ages of the
people attending the showing of a movie.
60
55
20
5
15
10 10
5 2
a. How many people attended this move?
5 + 60 + 55 + 20 + 10 + 15 + 10 + 5 + 2 =
182
2) The histogram below shows the ages of the
people attending the showing of a movie.
60
55
20
5
15
10 10
5 2
b. Circle the interval below that contains the mean age
of the people attending this movie. Explain how you
can make this choice without actually computing the
mean.
5-10 years old
15-20 years old
30-35 years old
2) HOW TO COMPUTE MEAN FROM HISTOGRAM:
60
55
20
10
5
2.5
7.5
12.5 17.5
15
10
32.5 37.5
42.5
1) Find the central value of each bin (interval).
5 2
67.5
72.5
2) HOW TO COMPUTE MEAN FROM HISTOGRAM:
60
55
20
10
5
2.5
7.5
12.5 17.5
15
10
32.5 37.5
42.5
5 2
67.5
2) Multiply central value times the entries in that bin
and add the products:
2.5(5) + 7.5(60) + 12.5(55) + 17.5(20) + 32.5(10) +
37.5(15) +42.5(10) + 67.5(5) + 72.5(2) = 2945
72.5
2) HOW TO COMPUTE MEAN FROM HISTOGRAM:
60
55
Mean age: 16.18
20
10
5
2.5
7.5
12.5 17.5
15
10
32.5 37.5
5 2
67.5
42.5
3) Divide this sum by total entries:
2945
 16.18
182
This means that the mean age is
16.18 which falls beteen 15-20.
72.5
2) The histogram below shows the ages of the
people attending the showing of a movie.
60
55
20
5
101510
5 2
c. Approximately, what is the median?
Since 182 people attended, the median number of people is (182+1) / 2 =
183 / = 91.5. Since 5 people are in the first bin, 60 people are in the
2
second bin (5 + 60 = 65), then the 91.5 person has to lie in the third bin
about ½ way through. This is the 10-15 age range. 12.5 is the age
about ½ way into this 10-15 interval.
2) The histogram below shows the ages of the
people attending the showing of a movie.
60
55
20
5
101510
d. Approximately, what is the mean?
2945
 16.18 or about 17
182
5 2
2) The histogram below shows the ages of the
people attending the showing of a movie.
55
26.5
28.5
41.25
60
5
Median age: 12.5
20
101510
5 2
Q1:7 12.5
e. Draw an approximate box-whiskers plot for the histogram.
Compute Q1: Find the number of people halfway from the
minimum to the median (91.5 + 1) / 2 = 46.25
Counting over from the minium, Q1 should fall around 7 or 8.
2) The histogram below shows the ages of the
people attending the showing of a movie.
55
26.5
28.5
41.25
60
Median age: 12.5
20
5
101510
5 2
Q1:7 12.5 Q3:7
e. Draw an approximate box-whiskers plot for the histogram.
Compute Q3: Find the number of people halfway from the
median to the maximum (91.5 + 1) / 2 = 46.25
Counting over from the median, Q3 should fall around 19 or 20.
2) The histogram below shows the ages of the
people attending the showing of a movie.
55
26.5
28.5
41.25
60
Median age: 12.5
20
5
min
Q1:7 12.5 Q3:7
101510
5 2
max
e. Draw an approximate box-whiskers plot for the histogram.
All values for the box and whisker plot are now labeled in
the histogram.
Min:
0
Max:
75
Q 1:
7
Q 2:
Q 3:
20
2) The histogram below shows the ages of the
people attending the showing of a movie.
f. Describe the spread in the data.
Skewed right
3)
.4
1.7 2.6 4
IQR = Q3 – Q1
=
4 – 1.7 = 2.3
20.1
(a.) Are there any outliers? Explain how you know.
HOW TO DETERMINE IF THERE IS(ARE) OUTLIERS:
If any value lies beyond either of the following, then these are
outliers. If no value lies beyond either of the following, then there
are no outliers.
Q1 – IQR(1.5) = 1.7 – 2.3(1.5) = 1.7 – 3.45 = - 1.75
No outlier to the left of Q1
Q3 + IQR(1.5) = 4 + 2.3(1.5) = 4 + 3.45 = 7.45
Since 20.1 lies to the right of 7.45, 20.1 is an outlier.
3) The box plot below shows the number of licensed drivers (in
millions) for 25 states in the United States. The minimum value
is 0.4 million, the lower quartile is 1.7 million, the median is 2.6
million, the upper quartile is 4.0 million, and the maximum is
20.1 million.
(b.) Describe the shape of the distribution. Explain your
reasoning.
Skewed right
3) The box plot below shows the number of licensed drivers (in
millions) for 25 states in the United States. The minimum value
is 0.4 million, the lower quartile is 1.7 million, the median is 2.6
million, the upper quartile is 4.0 million, and the maximum is
20.1 million.
(c.) How many of these states have more than 4 million
licensed drivers? Explain how you know.
25 % of the values lie above Q3.
So 25(.25) = 6.25 states
3)
(d.) Draw an appropriate histogram that matches this box and
whisker plot.
4) Tamika kept track of the number of minutes that she
exercised each day for two months. The distribution
was approximately normal. She calculated the
following summary statistics.
Mean = 35 minutes, median = 34 minutes,
standard deviation = 8 minutes, lower quartile = 24
minutes, and upper quartile = 41 minutes
(a.) Find the interquartile range and write a sentence
that describes what it tells you about Tamika’s
exercise times.
IQR = Q3 – Q1 = 41 – 24 = 17
50 % of her exercise minutes
fell within this range
4) Tamika kept track of the number of minutes that she
exercised each day for two months. The distribution
was approximately normal. She calculated the
following summary statistics.
Mean = 35 minutes, median = 34 minutes,
standard deviation = 8 minutes, lower quartile = 24
minutes, and upper quartile = 41 minutes
(b.) Suppose that she ran five more minutes each day
for the next two months. What would the new
quartiles be?
Mean: 40
Median: 39
Standard Deviation:
8
IQR: 17
5) Jack’s homework grades are 75, 78, 96, 94, 93, 100,
101, 93, 60, 65, 64, 62, 73, 85, 87.
Find the:
60, 62, 64, 65, 73, 75, 78, 85, 87, 93, 93, 94, 96, 100, 101
a.) mean 81.73
b.) median 85
c.) mode 93
d.) range 101 – 60 = 41
e.) IQR
Q3 – Q1 = 94 – 65 = 29
5) Jack’s homework grades are 75, 78, 96, 94, 93, 100,
101, 93, 60, 65, 64, 62, 73, 85, 87.
Find the:
60, 62, 64, 65, 73, 75, 78, 85, 87, 93, 94, 96, 100, 101
f.) standard deviation
14.49 or 14.006
5) Jack’s homework grades are 75, 78, 96, 94, 93, 100,
101, 93, 60, 65, 64, 62, 73, 85, 87.
Make a histogram and a box plot that displays his
homework grades.
max
101
min
60
Q1
65
60
Q2 Q3
85 94
70
80
90
100
110
6) The box plots below show the ages of the members
of the 2005 U. S. Olympic Hockey teams. Answer the
questions that follow using the information in these
plots. Be sure to clearly explain each of your
answers.
Min
18
Q1
22
Q2
24
Min
24
Q3
26
Q1
27
Max
31
Q2
32
Q3
34
a. Which team had the greater range in ages?
Women: 31 – 18 = 13
Men
Men: 43 – 24 = 19
Max
43
6) The box plots below show the ages of the members
of the 2005 U. S. Olympic Hockey teams. Answer the
questions that follow using the information in these
plots. Be sure to clearly explain each of your
answers.
Min
18
Q1
22
Q2
24
Min
24
Q3
26
Q1
27
Max
31
Q2
32
Q3
34
Max
43
b. What percentage of the female players were at least
22 years old?
75 %
6) The box plots below show the ages of the members
of the 2005 U. S. Olympic Hockey teams. Answer the
questions that follow using the information in these
plots. Be sure to clearly explain each of your
answers.
Min
18
Q1
22
Q2
24
Min
24
Q3
26
Max
31
Q1
27
Q2
32
Q3
34
Max
43
c. True or False? More men were older than 34 than
were younger than 27.
Both are in the
25 % range.
6) The box plots below show the ages of the members
of the 2005 U. S. Olympic Hockey teams. Answer the
questions that follow using the information in these
plots. Be sure to clearly explain each of your
answers.
Min
18
Q1
22
Q2
24
Min
24
Q3
26
Q1
27
Max
31
Q2
32
Q3
34
d. Which team had the higher standard deviation?
Men
Max
43
6) e. Draw a histogram for the men.
Min
18
Q1
22
Q2
24
Min
24
Q3
26
Q1
27
Max
31
Q2
32
Q3
34
Max
43
7) a. Which of the following examples are most likely
to have the largest interval ranges for a histogram
and interquartile range for a box plot? I
I: The weight (in pounds) of 30 people at the mall.
II: The ages of 30 people at the mall.
III: The amount of money in the wallet of 30 people at
the mall.
IV: The shoe sizes of 30 people at the mall.
b. Which of the above is most likely to have the
smallest intervals and quartile range?
IV
8) The following are attendance for home football
games: 287, 263, 286, 286, 256, 70, 255, and 300. If
the attendance for the 6th game (70) wasn’t included,
what would be the answer for each of the following?
a. Which measure would increase the most:
mean, mode, median, or standard deviation?
mean
b. Will the interquartile range increase or decrease?
neither
8) The following are attendance for home football
games: 287, 263, 286, 286, 256, 70, 255, and 300. If
the attendance for the 6th game (70) wasn’t included,
what would be the answer for each of the following?
c. Will the range increase or decrease?
decrease
d. Will the standard deviation increase or decrease?
decrease
9) A teacher interviewed 200 students and found the
following results:
Sophomore
Male
.2 .2(200) = 40
Female
.1 .1(200) = 20
Junior
Senior
.1 .1(200) = 20
.25 .25(200) = 50
.25 .25(200) = 50
.1 .1(200) = 20
a. How many senior males did the teacher interview?
50
b. If the student was a sophomore, what gender would
this student more likely be?
male
c. How many more female juniors were there than
male juniors?
50 – 20 = 30
10) A van holds 15 people. Which of the following
would you need to know if you wanted to calculate
the approximate total weight of the people in the
van? (median, mode, standard deviation, range, or
mean)
mean
11) Solve the following equations:
a.
b.
 5 x  20
-5 -5
x<-4
2 x  3  15
-3 -3
2 x = - 18
2
2
x=-9
11) Solve the following equations:
c.
14 x  7
14 14
x = 7 / 14 = ½ = .5
2 x  3  5 x  10
-2x
-2x
3 = 3 x – 10
+ 10
+ 10
e. 5  x  12
-5
-5
d.
13 = 3 x
3
3
x = 13 / 3 = 4.3
- x = - 17
(- 1)(- x) = (- 17)(- 1)
x = 17