REVIEW OF UNIT 1

REVIEW OF UNIT 1

1) The table displays the number of videos rented.

Number of Videos Rented

Number of Families

3 164 146 118 210 4

a. How many families were surveyed?16 + 14 + 11 + 2 +

4 = 47

REVIEW OF UNIT 1



Number of Families

3 164 146 118 210 4

b. Find the median number of videos rented by these families.

Median # of families: 47+1 / 2= 48 / 2 = 24Since 16 families rented 3 videos, the 24 families would fall into the 4- video category. 16 + 14 = 30

REVIEW OF UNIT 1



Number of Families

3 164 146 118 210 4

c. Find the mean number of videos rented by these families.

81.447226

47)4(10)2(8)11(6)14(4)16(3

2) The histogram below shows the ages of the people attending the showing of a movie.

a. How many people attended this move?

5

6055

20101510 5 2

5 + 60 + 55 + 20 + 10 + 15 + 10 + 5 + 2 = 182


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

5

6055

20101510 5 2

2) HOW TO COMPUTE MEAN FROM HISTOGRAM:

5

6055

2010

1510 5 2

1) Find the central value of each bin (interval).

2.5 7.512.5

17.5

32.5

37.5

42.5

67.5 72.

5


5

6055

2010

1510 5 2

2) Multiply central value times the entries in that bin and add the products:

2.5 7.512.5

17.5

32.5

37.5

42.5

67.5 72.

5

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


5

6055

2010

1510 5 2

3) Divide this sum by total entries:

2.5 7.512.5

17.5

32.5

37.5

42.5

67.5 72.

5

18.161822945

This means that the mean age is 16.18 which falls beteen 15-20.

Mean age: 16.18


c. Approximately, what is the median?

5

6055

20101510 5 2

Since 182 people attended, the median number of people is (182+1) / 2 = 183 / 2 = 91.5. Since 5 people are in the first bin, 60 people are in the 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.


d. Approximately, what is the mean?

1718.161822945

aboutor

5

6055

20101510 5 2


e. Draw an approximate box-whiskers plot for the histogram.

5

6055

20101510 5 2

12.5

Median age: 12.526

.5

28.

5

Compute Q1: Find the number of people halfway from the minimum to the median (91.5 + 1) / 2 = 46.25

41.2

5

Counting over from the minium, Q1 should fall around 7 or 8.

Q1:7



5

6055

20101510 5 2

12.5

Median age: 12.526

.5

28.

5

Compute Q3: Find the number of people halfway from the median to the maximum (91.5 + 1) / 2 = 46.25

41.2

5

Counting over from the median, Q3 should fall around 19 or 20.

Q1:7 Q3:7



5

6055

20101510 5 2

12.5

Median age: 12.526

.5

28.

5

All values for the box and whisker plot are now labeled in the histogram.

41.2

5

Q1:7 Q3:7min max

Q2: 12.5

Q1: 7

Q3: 20

Min: 0

Max: 75


f. Describe the spread in the data.

Skewed right

3)

(a.) Are there any outliers? Explain how you know.

.41.72.6 4 20.1

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

IQR = Q3 – Q1 = 4 – 1.7 = 2.3

No outlier to the left of Q1Q3 + IQR(1.5) = 4 + 2.3(1.5) = 4 + 3.45

= 7.45Since 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

(d.) Draw an appropriate histogram that matches this box and whisker plot.

3)

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 = 1750 % 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: Median:

Standard Deviation: IQR:

40

39

8 17

5)Jack’s homework grades are 75, 78, 96, 94, 93, 100, 101, 93, 60, 65, 64, 62, 73, 85, 87.

Find the:

a.) mean

b.) medianc.) mode

d.) range

e.) IQR

60, 62, 64, 65, 73, 75, 78, 85, 87, 93, 93, 94, 96, 100, 101

81.73

85

93

101 – 60 = 41

Q3 – Q1 = 94 – 65 = 29


Find the:

f.) standard deviation

60, 62, 64, 65, 73, 75, 78, 85, 87, 93, 94, 96, 100, 101

14.49 or 14.006


Make a histogram and a box plot that displays his homework grades.

60 70 80 90 100 110

Q1 65

Q2 85

Q3 94

min 60

max 101

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.

a. Which team had the greater range in ages?

Min 18

Max 31

Q1 22

Q2 24

Q3 26

Min 24

Max 43

Q1 27

Q2 32

Q3 34

Women: 31 – 18 = 13 Men: 43 – 24 = 19

Men



b. What percentage of the female players were at least 22 years old?

Min 18

Max 31

Q1 22

Q2 24

Q3 26

Min 24

Max 43

Q1 27

Q2 32

Q3 34

75 %



c. True or False? More men were older than 34 than were younger than 27.

Min 18

Max 31

Q1 22

Q2 24

Q3 26

Min 24

Max 43

Q1 27

Q2 32

Q3 34

Both are in the 25 % range.



d. Which team had the higher standard deviation?

Min 18

Max 31

Q1 22

Q2 24

Q3 26

Min 24

Max 43

Q1 27

Q2 32

Q3 34

Men

6) e. Draw a histogram for the men.

Min 18

Max 31

Q1 22

Q2 24

Q3 26

Min 24

Max 43

Q1 27

Q2 32

Q3 34

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: 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?

I

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?

b. Will the interquartile range increase or decrease?

mean

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?

d. Will the standard deviation increase or decrease?

decrease

decrease

9) A teacher interviewed 200 students and found the following results:

Male FemaleSophomore .2 .1Junior .1 .25Senior .25 .1

a. How many senior males did the teacher interview?

b. If the student was a sophomore, what gender would this student more likely be?

c. How many more female juniors were there than male juniors?

.2(200) = 40.1(200) = 20.25(200) = 50

.1(200) = 20.25(200) = 50.1(200) = 20

50

male

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. 205 x

b.

1532 x

- 5

- 5x < -

4

- 3

- 32 x = -

182 2x = - 9

11) Solve the following equations:

c.

10532 xxd.

714 x

125 xe.

14 14 x = 7 / 14 = ½ = .5

- 2 x

- 2 x3 = 3 x –

10 + 10

+ 10

13 = 3 x 33

x = 13 / 3 = 4.3

- 5 - 5 - x = - 17(- 1)(- x) = (- 17)(- 1) x = 17

REVIEW OF UNIT 1

Documents

Transcript of REVIEW OF UNIT 1