Unformatted text preview: is
If
skewed to the right
skewed
(i.e. positively skewed), as the left
positively
),
part is more dense and the right
tail is longer (Figure 2).
• • When Mean< Median, the
When
distribution is skewed to the left
skewed
(i.e. negatively skewed), as the
negatively
),
right part is more dense and the
left tail is longer (Figure 3).
left The formula for Skewness is :
The 3( µ − Med )
Sk =
σ Example 5:
Find the skewness for the data of Table 1.
Solution: 3(45.75 − 47.45)
Sk =
= −0.3687 = −36.87%
13.8333
(See Ex 2 & Ex 3 for the calculation of the mean, s.d. and
median)
median) Some interesting examples
Some
For the table, state the
For
boundaries of the class
interval if the data
referred to are:
(i) lengths
(i)
(ii) ages Class Interval frequency 1519 11 2024 27 2529 38 3034 41 ….. …… ….. ….. Solution
Solution
(i) For lengths the boundaries are
lengths
shown in the table:
shown For cases such as measuring we
For
employ the usual rounding up
and down practice.
and
For example,
For measurement data values such as
For
19.5, 19.6, 19.7, 19.8, 19.9 we
round up to 20.
For data such as 24.0, 24.1, 24.2,
24.3, 24.4 we round down to
24.
24.
So data in the range of 19.5 to 24.5
So
belong in the class 2024.
belong Class Interval Class
Class
boundary
boundary 1519 14.519.5 2024 19.524.5 2529 24.529.5 3034 29.534.5 ….. …… ….. ….. Although our true age may be 24years
Although
1day, 24yrs 2months, 24years
11mths, 24years 11mnths 29days,
we still give our age as 24.
we
A person is 24 until the day of their
person
birthday and then their 25.
birthday
This idea involves TRUNCATION as
This
the method of rounding.
the
Eg. Consider the numbers 20.0, 20.1,
Eg.
20.2, 20.3, 20.4, 20.5, 20.6, 20.7,
20.8, 20.9, 20.99, 20.99999
when truncated they all become
20.
20.
Now, consider the numbers 24.0, 24.1,
Now,
24.2, 24.3, 24.4, 24.5, 24.6, 24.7,
24.8, 24.9, 24.99, 24.99999
when truncated they all become
24.
24.
So data in the range of 20.0 to 24.9999
So
belong in the class 2024.
belong Solution
Solution
Class Interval Class
Class
boundary
boundary 1519 1520 2024 2025 2529 2530 3034 305 ….. …… ….. ….. Compare these five examples:
Compare
Question 1. (See Simple Exercises Ch4 , Q5) Find the mode, median, range, mean, s.d. and variance of the following data:
Find
of 4, 32, 2 , 4, 7, 8
4,
Solution: Order the data:
Order This is UNGROUPED data
This UNGROUPED
Mode = 4
Median = 5.5
Range = 30
Mean = 9.5
Mean
s.d. = 10.259
Variance = 105.25 2, 4, 4, 7, 8, 32
2,
N = 6 (even number of data). Note: “32” is an “extreme” value.
Note
Here, the median is a better indicator of the centre than the mean. Q2:
Q2:
Find the mean, s.d., variance, mode, median, range. x 6.2 7.5 8.3 9.1 total
f 10 15 20 25
N=70
N=70
This is UNGROUPED data.
This UNGROUPED Mean = 8.114
s.d. = 0.9804
Variance = 0.9612
Mode = 9.1
Median = 8.3
Range = 2.9 It’s MULTIPLE data.
MULTIPLE Q3:
Q3:
Find the mean, s.d., variance, mode, MEDIAN, range. x...
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 Fall '13
 Sets, Standard Deviation, Frequency distribution

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