Measures of Spread & Standard Deviation
SECONDARY 1 MATH
Distribution
overall shape of a set of data in a representation
Box & Whisker Histogram
Bell Curve
Normal
mean & median are close together
(same interval in histogram)
Box & Whisker: distance from Min to Med is about the same as from Med to Max
mean is lower than the median
(different intervals in histogram)
Skew Left
data is spread to the left, concentrated on the right
Skew Right
mean is higher than the median
data is spread to the right, concentrated on the left
(different intervals in histogram)
Symbol for Mean
Symbol for Standard Deviation
“x bar”
Greek letter: “mu”
Greek letter: “sigma”
Mark the mean and median in the interval where they occur
Med
Mean: 81.14 Median: 82
Determine if the distribution is normal, skew left, or skew right
Med
Normal Distribution
both the mean and median are close together,
they are in the same interval
Label the titles of each axis
Med
Frequency
Math Test Scores
Finish labeling the histogram on both axes and titles
30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109
Frequency
Math Test Scores
12
11
10
9
8
7
6
5
4
3
2
1
Mark the median and mean in the interval where they occur
30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109
Frequency
Math Test Scores
12
11
10
9
8
7
6
5
4
3
2
1
Med
Median: 82 Mean: 79.66
30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109
Frequency
Math Test Scores
12
11
10
9
8
7
6
5
4
3
2
1
Med
Determine if the distribution is normal, skew left, or skew right
Slightly Skewed Left
the mean is lower than the median and
they are in different intervals
the mean (average) was pulled lower because of the new data added
Finish labeling the histogram on both axes and titles
Frequency
Math Test Scores
10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109
18
16
14
12
10
8
6
4
2
Frequency
Math Test Scores
10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109
18
16
14
12
10
8
6
4
2
Mark the median and mean in the interval where they occur
Med
Median: 15 Mean: 20.11
Frequency
Math Test Scores
10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109
18
16
14
12
10
8
6
4
2
Med
Determine if the distribution is normal, skew left, or skew right
Skewed Right
mean is higher than the median and in different intervals
the very high scores pulled the data shape upward (to the right)
Measures of Spread
A statistic that tells you how dispersed, or spread out, data values are.
Standard Deviation: A number that shows how much variation
or “dispersion” there is from the mean. A low standard of deviation
indicates that the data points tend to be very close to the mean.
A high standard of deviation indicates that the data are spread out over a large range of values.
sigma
(standard deviation)
mean
number of data points
each data point
68.2% of the data is within one standard deviation of the mean
95.4% of the data is within two standard deviations of the mean
Standard Deviation in
a Normal Distribution
Office A Office B
14, 17, 18, 19, 20, 8, 11, 12, 16, 18
24, 24, 30, 32 18, 18, 20, 23
Find the standard deviation for the waiting times in each data set
Find the standard deviation for Office A using the formula
Find the mean:
(14 – 22)2
(17 – 22)2
(18 – 22)2
(19 – 22)2
…
Office A Office B
14, 17, 18, 19, 20, 8, 11, 12, 16, 18
24, 24, 30, 32 18, 18, 20, 23
Find the standard deviation for the waiting times in each data set
Calculate one standard deviation above & below the mean
22 – 5.68 =
22 + 5.68 =
16.32 minutes
27.68 minutes
Most waiting times are within 5.68 minutes of the average wait time of 22 minutes.
of the numbers were within 1 standard deviation of the mean 67%
Most waiting times are between 16.32 minutes and 27.68 minutes
Office A Office B
14, 17, 18, 19, 20, 8, 11, 12, 16, 18
24, 24, 30, 32 18, 18, 20, 23
Find the standard deviation for the waiting times in each data set
Find the standard deviation for Office B
using a graphing calculator
Office A Office B
14, 17, 18, 19, 20, 8, 11, 12, 16, 18
24, 24, 30, 32 18, 18, 20, 23
Find the standard deviation for the waiting times in each data set
Find the standard deviation for Office B
using a graphing calculator
Office A Office B
14, 17, 18, 19, 20, 8, 11, 12, 16, 18
24, 24, 30, 32 18, 18, 20, 23
Find the standard deviation for the waiting times in each data set
Calculate one standard deviation above & below the mean
16 – 4.50 =
16 + 4.50 =
11.50 minutes
20.50 minutes
Most waiting times are within 4.50 minutes of the average wait time of 16 minutes.
67% again
Most waiting times are between 11.50 minutes and 20.50 minutes at Office B