Descriptive statistics. Elementary statistics. Larson. Farber. (Chapter 2)

97 107 67 78 125
109 99 105 99 101 92

Make a frequency distribution table with five classes.

Minutes Spent on the Phone

Key values:

Minimum value =
Maximum value =

67

125

Calculate the Class Width
(125 - 67) / 5 = 11.6 Round up to 12
Determine Class Limits
Mark a tally in appropriate class for each data value

Frequency Distributions

78
90
102
114
126

3
5
8
9
5

67
79
91
103
115

Do all lower class limits first.

-126

3
5
8
9
5

Midpoint: (lower limit + upper limit) / 2

Relative frequency: class frequency/total frequency

Cumulative frequency: Number of values in that class or in
lower one.

Other Information

Midpoint

Relative
frequency

Cumulative
frequency

72.5
84.5
96.5
108.5
120.5

0.10
0.17
0.27
0.30
0.17

3
8
16
25
30

-114.5
115.5 -126.5

Frequency Histogram

Time on Phone

minutes

f

each bar. Connect consecutive midpoints. Extend the frequency polygon to the axis.

that
are less than or equal to the given value, x.

and highest value is 125, so list stems from 6 to 12.

102 124 108 86 103 82

2

4

8

6

3

2

7 7
9 |2 5 7 9 9
10 |0 1 2 3 3 4 5 5 7 8 9
11 |2 6 8
12 |2 4 5

Key: 6 | 7 means 67

1
7 | 8
8 | 2
8 | 5 6 7 7
9 | 2
9 | 5 7 9 9
10 | 0 1 2 3 3 4
10 | 5 5 7 8 9
11 | 2
11 | 6 8
12 |2 4
12 | 5

Key: 6 | 7 means 67

1st line digits 0 1 2 3 4

2nd line digits 5 6 7 8 9

1st line digits 0 1 2 3 4

2nd line digits 5 6 7 8 9

Chart

Used to describe parts of a whole
Central Angle for each segment

Construct a pie chart for the data.

by the number of values
For a population: For a sample:

Median: The point at which an equal number of values fall above and fall below

Mode: The value with the highest frequency

the median, and the mode

n = 9

Mean:

Median: Sort data in order

0 2 2 2 3 4 4 6 40

The middle value is 3, so the median is 3.

Mode: The mode is 2 since it occurs the most times.

An instructor recorded the average number of absences for his students in one semester. For a random sample the data are:

median, and the mode

n =8

Mean:

Median: Sort data in order

The middle values are 2 and 3, so the median is 2.5

Mode: The mode is 2 since it occurs the most.

Suppose the student with 40 absences is dropped from the course. Calculate the mean, median and mode of the remaining values. Compare the effect of the change to each type of average.

0 2 2 2 3 4 4 6

Fridays. Calculate the mean, median and mode for each.

Mean = 61.5
Median =62
Mode= 67

Mean = 61.5
Median =62
Mode= 67

56 33
56 42
57 48
58 52
61 57
63 67
63 67
67 77
67 82
67 90

Stock A

Stock B

value - Minimum value

Range for B = 90 - 33 = $57

The range only uses 2 numbers from a data set.

The deviation for each value x is the difference between the value of x and the mean of the data set.

In a population, the deviation for each value x is:x - μ

In a sample, the deviation for each value x is:

Measures of Variation

1.5
5.5
5.5
5.5

56
56
57
58
61
63
63
67 67 67

Deviations

µ = 61.5

56 - 61.5

56 - 61.5

57 - 61.5

58 - 61.5

∑ ( x - µ) = 0

Stock A

Deviation

The sum of the deviations is always zero.

by N.

Stock A
56 -5.5 30.25
56 -5.5 30.25
57 -4.5 20.25
58 -3.5 12.25
61 -0.5 0.25
63 1.5 2.25
63 1.5 2.25
67 5.5 30.25
67 5.5 30.25
67 5.5 30.25

188.50

Sum of squares

Population Variance

population variance.

The population standard deviation is $4.34

calculate a sample variance divide the sum of squares by n-1.

The sample standard deviation, s is found by taking the
square root of the sample variance.

value

Population Variance

the following characteristics.

About 68% of the data lies within 1 standard deviation of the mean

About 99.7% of the data lies within 3 standard deviations of the mean

About 95% of the data lies within 2 standard deviations of the mean

68%

is $125 thousand with a standard deviation of $5 thousand. The data set has a bell shaped distribution. Estimate the percent of homes between $120 and $135 thousand

68%

68%

$120 is 1 standard deviation below the mean and $135 thousand is 2 standard deviation above the mean.

68% + 13.5% = 81.5%

So, 81.5% of the homes have a value between $120 and $135 thousand .

68%

of the data lies within 3 standard deviation of the mean.

For any distribution regardless of shape the portion of data lying within k standard deviations (k >1) of the mean is at least 1 - 1/k2.

μ =6
σ =3.84

For k = 2, at least 1-1/4 = 3/4 or 75% of the data lies within 2 standard deviation of the mean.

seconds with a standard deviation of 2.2 sec. Apply Chebychev’s theorem for k = 2.

52.4

54.6

56.8

59

50.2

48

45.8

2 standard deviations

At least 75% of the women’s 400- meter dash times will fall between 48 and 56.8 seconds.

Mark a number line in standard deviation units.

a frequency distribution, treat each value as if it occurs at the midpoint
of its class. x = Class midpoint.

x f

2991

distribution,
use x = class midpoint.

739.84

2219.52

231.04

1155.20

10.24

81.92

77.44

696.96

432.64

2163.2

30

6316.8

27 randomly selected days in the last year is given. Find Q1, Q2 and Q3..
28 43 48 51 43 30 55 44 48 33 45 37 37 42 27 47 42 23 46 39 20 45 38 19 17 35 45

3 quartiles Q1, Q2 and Q3 divide the data into 4 equal parts.
Q2 is the same as the median.
Q1 is the median of the data below Q2
Q3 is the median of the data above Q2

23 27 28 30 33 35 37 37 38 39 42 42
43 43 44 45 45 45 46 47 48 48 51 55 .

Quartiles

Median rank (27 +1)/2 = 14. The median = Q2 = 42.

There are 13 values below the median.
Q1 rank= 7. Q1 is 30.
Q3 is rank 7 counting from the last value. Q3 is 45.

The Interquartile Range is Q3 - Q1 = 45 - 30 = 15

values to describe a set of data. Q1, Q2 and Q3, the minimum value and the maximum value.

Q1
Q2 = the median
Q3
Minimum value
Maximum value

30
42
45
17
55

Interquartile Range

P1, P2, P3…P99 .

A 63nd percentile score indicates that score is greater than or equal to 63% of the scores and less than or equal to 37% of the scores.

P50 = Q2 = the median

P25 = Q1

P75 = Q3