Transcript Bimodal Distribution

LESSON 11–1
Descriptive Statistics
Five-Minute Check (over Chapter 10)
Then/Now
New Vocabulary
Key Concept: Symmetric and Skewed Distributions
Key Concept: Choosing Summary Statistics
Example 1: Skewed Distribution
Example 2: Bimodal Distribution
Key Concept: Symmetric and Skewed Box Plots
Example 3: Describe a Distribution Using a Box Plot
Example 4: Compare Position Using Box Plots
Example 5: Construct and Use a Percentile Graph
Over Chapter 10
Find the fifth term of the sequence
A.
B.
C.
D.
Over Chapter 10
Write an explicit formula and a recursive formula
for finding the nth term of the arithmetic sequence
−2, 5, 12, 19, … .
A. an = 7n – 3; a1 = –2, an = an – 1 + 3
B. an = –2n + 7; a1 = –2, an = 7an – 1 – 2
C. an = –2 + 7n; a1 = –2, an = –2 + 7n
D. an = 7n – 9; a1 = –2, an = an – 1 + 7
Over Chapter 10
Evaluate
A. 181,398,527
B. 60,455,176
C. 177,146
D. 118,098
Over Chapter 10
Use Pascal’s triangle to expand (x – 2y)4.
A. x 4 – 8x 3y + 24x 2y 2 – 32xy 3 + 16y 4
B. x 4 + 4x 3y + 6x 2y 2 + 4xy 3 + 1
C. x 4 – 8x 3y + 12x 2y 2 – 8xy 3 + 2y 4
D. x 4 + 8x 3y + 24x 2y + 32xy 3 – 16y 4
Over Chapter 10
Which of the following represents −4 − 4i in
exponential form?
A.
B.
C.
D.
You found measures of central tendency and standard
deviations. (Lesson 0-8)
• Identify the shapes of distributions in order to select
more appropriate statistics.
• Use measures of position to compare two sets of
data.
• univariate
• percentile
• negatively skewed distribution
• percentile graph
• symmetrical distribution
• positively skewed distribution
• resistant statistic
• cluster
• bimodal distribution
Skewed Distribution
A. GOLF Twenty members of a golf team each hit
a golf ball as far as they could. The table shows
the distance each team member drove his or her
ball. Construct a histogram and use it to describe
the shape of the distribution.
Skewed Distribution
On a graphing calculator, press STAT EDIT and input
the data into L1. Then turn on Plot1 under the STAT
PLOT menu and choose
. Graph the histogram by
pressing ZoomStat or by pressing Graph and
adjusting the window manually.
The graph shown has three equal bars. Using the
TRACE feature, you can determine that these peaks
represent distances from 185 to 215 yards.
Skewed Distribution
Answer: The graph is positively skewed. Most golf
shots range from 185 to 215 yd, but a few
shots were much longer, so the tail of the
distribution trails off to the right.
Skewed Distribution
B. GOLF Twenty members of a golf team each hit
a golf ball as far as they could. The table shows
the distance each team member drove his or her
ball. Summarize the center and spread of the data
using either the mean and standard deviation or
the five-number summary. Justify your choice.
Skewed Distribution
Since the distribution is skewed, use the five-number
summary instead of the mean and standard deviation
to summarize the center and spread of the data. To
display this summary, press STAT, select 1-Var Stats
under the Calc submenu, and scroll down.
Answer: five-number summary; The median is 205
yd, and the interquartile range is from 195
to 212.5 yd.
TRAVEL The distance from Reemi's
lake cottage to 20 towns are given in
the table. Summarize the center and
spread of the data using either the
mean and standard deviation or the
five-number summary. Justify your
choice.
A.
mean 62.8, standard deviation 23.35; because the graph is
symmetrical
B.
mean 62.8, standard deviation 23.35; because the graph is
skewed
C.
median 67, interquartile range from 50.5 to 82.5; because
the graph is positively skewed
D.
median 67, interquartile range from 50.5 to 82.5; because
the graph is negatively skewed
Bimodal Distribution
A. CELL PHONES People from two professions
were asked how much they paid for their work cell
phone. Construct a histogram and use it to
describe the shape of the distribution.
Bimodal Distribution
The histogram of the data has not one but two major
peaks. Therefore, the distribution is neither
symmetrical nor skewed but bimodal. The two
separate clusters suggest that two types of cell
phones are mixed in the data set. It is likely that there
are several relatively inexpensive cell phones and
some much more expensive ones.
Bimodal Distribution
Answer: The distribution is bimodal.
Bimodal Distribution
B. CELL PHONES People from two professions
were asked how much they paid for their work cell
phone. Summarize the center and spread of the
data using either the mean and standard deviation
or the five-number summary. Justify your choice.
Bimodal Distribution
Since the distribution is bimodal, an overall summary
of center and spread would give an inaccurate
depiction of the data. Instead, summarize the center
and spread of each cluster. Since the lower cluster of
the data is positively skewed, the five-number
summary shows that the prices ranged from $50 to
$70, with a median of $60, and half of the prices were
between $55 and $60. Since the upper cluster is
symmetrical, the mean of $100 and standard deviation
of $4.50 can be used to describe the center and
spread, respectively.
Bimodal Distribution
Bimodal Distribution
Answer: Sample answer: The lower cluster of the
data is positively skewed, so the fivenumber summary shows that the prices
ranged from $50 to $70, with a median of
$60, and half of the prices were between
$55 and $60. Since the upper cluster is
symmetrical, the mean of $100 and
standard deviation of $4.50 can be used to
describe the center and spread,
respectively.
HOTEL PRICES Summarize the center and spread
of the data using either the mean and standard
deviation or the five-number summary. Justify
your choice.
A.
The distribution is bimodal. Since the lower cluster is positively
skewed, the five-number summary indicates that the prices
range from $49 to $99, with a median of $70.5, and half are
between $56 and $82. The upper cluster is symmetrical; the
mean price is $121.7 with standard deviation $16.6.
B.
The distribution is bimodal. Since the lower cluster is
symmetrical, the mean price is $71 with standard deviation
$15.5. The upper cluster is positively skewed so the fivenumber summary indicates that the prices range from $100 to
$150, with a median of $118, and half are between $108 and
$135.
C.
The distribution is symmetrical. The mean is $93.32 with
standard deviation $29.8.
D.
The distribution is negatively skewed. The five-number
summary shows that the prices ranged from $49 to $150, with
a median of $95, and half the prices between $69 and $115.
Describe a Distribution Using a Box Plot
A. HOME SALES The table shows the prices of
homes recently sold by a realtor. Construct a box
plot and use it to describe the shape of the
distribution.
Describe a Distribution Using a Box Plot
Input the data into L1 on a graphing calculator. Then
turn on Plot1 under the STAT PLOT menu and
choose . Graph the box plot by pressing ZoomStat
or by pressing Window and adjusting the window
manually.
Describe a Distribution Using a Box Plot
Since the left whisker is longer than the right whisker
and the line representing the median is closer to Q3
than to Q1, the distribution is negatively skewed.
Answer: The distribution is negatively skewed.
Describe a Distribution Using a Box Plot
B. HOME SALES The table shows the prices of
homes recently sold by a realtor. Summarize the
center and spread of the data using either the
mean and standard deviation or the five-number
summary. Justify your choice.
Describe a Distribution Using a Box Plot
Since the distribution is skewed, use the five-number
summary. This summary indicates that while the home
sale prices ranged from $95,000 to $184,000, the
median price was $165,000. Prices in the middle half
of the data varied by $170,000 – $150,000 or $20,000,
which is the interquartile range.
Describe a Distribution Using a Box Plot
Answer: Since the distribution is skewed, use the
five-number summary. The median is
$165,000. The interquartile range is
$20,000. The range is $89,000.
A. BOOKS The number of pages in the books that
Cecelia read this year is shown in the table.
Construct a box plot and use it to describe the
shape of the distribution.
A. The distribution
is symmetrical.
C. The distribution is
negatively skewed.
B. The distribution
is bimodal.
D. The distribution is
positively skewed.
Compare Position Using Box Plots
GAMES Jessica and Miguel earned these scores
on a new computer game. Construct side-by-side
box plots of the data sets. Then use this display to
compare the distributions.
Compare Position Using Box Plots
Input the data into L1 and L2. Then turn on Plot1 and
Plot2 under the STAT PLOT menu, chose
, and
graph the box plots by pressing ZoomStat or by
pressing Graph and adjusting the window manually.
Compare Position Using Box Plots
Miguel's median score of 84 was much higher than
Jessica's median score of 35. The first quartile in
Miguel's distribution of scores (81) is much higher than
Jessica's third quartile of 38, making the difference
substantial. Miguel's distribution is negatively skewed.
Jessica's is positively skewed.
Answer: Miguel's median score of 84 was much
higher than Jessica's median score of 35.
The first quartile in Miguel's distribution of
scores (81) is much higher than Jessica's
third quartile of 38, making the difference
substantial. Miguel's distribution is
negatively skewed. Jessica's is positively
skewed.
GROCERIES The amount the Dentons and the
Sherwoods spent on groceries each week are
shown in the tables. Construct side-by-side box
plots of the data sets to compare the distributions.
A.
The Sherwood's median is about equal to the
Denton's top expenditure. Half of the
Sherwood's amounts are more than any of the
Denton's.
B.
The Sherwood's median amount is greater than
the Denton's top amount. More than half of the
Sherwood's amounts are greater than all those
of the Dentons.
C.
The Sherwood's median amount is in the top
quartile of the Denton's amounts. Three-fourths
of the Denton's amounts are less than all of
those of the Sherwoods.
D.
The Sherwood's least amount is greater than the
third quartile of the Denton's amounts. More
than three-fourths of the Denton's amounts are
less than all those of the Sherwoods.
Construct and Use a Percentile Graph
A. FOOTBALL The table gives the frequency
distribution of weights of football players from 11
teams. Construct a percentile graph of the data.
Construct and Use a Percentile Graph
First, find the cumulative frequencies. Then find the
cumulative percentages by expressing the cumulative
frequencies as percents. The calculations for the first
two players are shown.
Construct and Use a Percentile Graph
Finally, graph the data with the class boundaries along
the x-axis and the cumulative percentages along the
y-axis, as shown.
Answer:
Construct and Use a Percentile Graph
B. FOOTBALL The table gives the frequency
distribution of weights of football players from 11
teams. Estimate the percentile rank a player
weighing 210 pounds would have in this
distribution and interpret its meaning.
Construct and Use a Percentile Graph
Find 210 on the x-axis and draw a vertical line to the
graph. This point on the graph corresponds to
approximately the 79th percentile. Therefore, a
football player weighing 210 pounds is heavier than
about 79% of the other players included in the
distribution.
Construct and Use a Percentile Graph
Answer: About 79th percentile; the football player is
heavier than about 79% of the other
players included in the distribution.
MOVIES The table gives the frequency
distribution of running time for 25 movies.
Estimate the percentile rank of a movie that runs
110 minutes.
A. 55%
B. 50%
C. 35%
D. 30%
LESSON 11–1
Descriptive Statistics