assessment of learning 2 (overview)
TRANSCRIPT
What is Frequency?โข Frequency is how often
something occurs.
Frequency Distribution - is a tabular arrangement of data into appropriate categories showing the number of observations in each category or group.
NASCAR Nextel Cup Driversโ Ages
Parts of a frequency distribution :โข Class limitโข Class sizeโข Class boundariesโข Class marks
Class limit- Is the groupings defined by
the lower and upper limits.Example: LL โ UL
10 โ 14 15 โ 19
20 โ 24
Lower class limit (LL)- smallest number in each
group
Upper class limit (UL)- highest number in each group
Class size- Is the width of each class
interval.Example: LL โ UL
10 โ 14 15 โ 19
20 โ 24 What is the class size of the
score distribution above? Answer: 5
Class boundaries- Are the numbers used to
separate each category in the distribution .
Example: LL โ UL 10 โ 14 9.5 โ 14.5 15 โ 19 14.5 โ 19.5
20 โ 24 19.5 โ 24.5
Class marks- Are the midpoint of the lower
and upper class limits.- Formula is =
Example: LL โ UL 10 โ 14 -
15 โ 19 - 20 โ 24 -
Class marks- Are the midpoint of the lower
and upper class limits.- Formula is =
Example: LL โ UL 10 โ 14 12
15 โ 19 17 20 โ 24 22
Steps in constructing Frequency Distribution
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Computing the Range
The range is the difference between the highest score and the lowest score.
Range = HS - LS
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Lowest score
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Lowest score
Highest score
Computing the RangeThe range is the difference
between the highest score and the lowest score.
Range = HS โ LS R= 50 โ 15 R= -
Computing the RangeThe range is the difference
between the highest score and the lowest score.
Range = HS โ LS R= 50 โ 15 R= 35
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Solve the value of k:k = 1 + 3.3 log n
Determine the class size (c.i).formula: or
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40
Determine the class size (c.i).formula: or
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 +
3.3(1.60205991)
Determine the class size (c.i).formula: or
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 +
3.3(1.60205991)k = 1 + 5.286797971
Determine the class size (c.i).formula: or
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 +
3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971
Determine the class size (c.i).formula: or
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 +
3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6
Determine the class size (c.i).formula: or
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 +
3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6
Determine the class size (c.i).formula: or
Find the class size:
c.i =
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 +
3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6
Determine the class size (c.i).formula: or
Find the class size:
c.i = c.i =
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 +
3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6
Determine the class size (c.i).formula: or
Find the class size:
c.i = c.i = c.i = 5.833
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 +
3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6
Determine the class size (c.i).formula: or
Find the class size:
c.i = c.i = c.i = 5.833c.i = 6
From Assessment of Learning 1, Yonardo Gabuyo
Solve the value of k:k =
Determine the class size (c.i).formula: or
Solve the value of k:k = k =
Determine the class size (c.i).formula: or
Solve the value of k:k = k = k = 6.32455532
Determine the class size (c.i).formula: or
Solve the value of k:k = k = k = 6.32455532k = 6
Determine the class size (c.i).formula: or
Find the class size:
c.i = c.i = c.i = 5.833c.i = 6
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Set up the class limits.X
15 โ 2021 โ 2627 โ 3233 โ 3839 โ 4445 - 50
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Make a tally X Tally
15 โ 20 / / / / /21 โ 26 / / / / / / / / /27 โ 32 / / / /33 โ 38 / / / / / / / / / /39 โ 44 / / / /45 - 50 / / / / / / / /
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Compute the
Range;R=HS-LS
Determine the class size.
Set up the class
limits.
Make a tally.
Count the
frequency (f).
Indicate the frequency. X Tally Frequency (f)
15 โ 20 / / / / / 521 โ 26 / / / / / / / / / 927 โ 32 / / / / 433 โ 38 / / / / / / / / / / 1039 โ 44 / / / / 445 - 50 / / / / / / / / 8
n = 40
Frequency distribution X Tally f cf
15 โ 20 / / / / / 5 17.5 521 โ 26 / / / / / / / / / 9 23.5 1427 โ 32 / / / / 4 29.5 1833 โ 38 / / / / / / / / / / 10 35.5 2839 โ 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40
n = 40
Graphical Representation of
Scores in Frequency Distribution
There are different methods of graphing frequency distribution:
โข Bar graphโข Histogramโข Frequency polygonโข Pie graph
Bar graphWhen data is in categories (countries, movies, music, etc.), this type of graph is usually used.
Rodents Fish Cats Dogs Rabbits0%
10%
20%
30%
40%
50%
60% Pet Preference
Pet Preferencepets
perc
enta
ge
Frequency distribution X Tally f cf
15 โ 20 / / / / / 5 17.5 521 โ 26 / / / / / / / / / 9 23.5 1427 โ 32 / / / / 4 29.5 1833 โ 38 / / / / / / / / / / 10 35.5 2839 โ 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40
n = 40
Frequency distribution X Tally f cf
15 โ 20 / / / / / 5 17.5 521 โ 26 / / / / / / / / / 9 23.5 1427 โ 32 / / / / 4 29.5 1833 โ 38 / / / / / / / / / / 10 35.5 2839 โ 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40
n = 40
15 - 20 21 - 26 27 - 32 33 - 38 39 - 44 45 -500
2
4
6
8
10
12
Scores
Scores
Classes
freq
uenc
y
HistogramGroups numbers into ranges and is a great way to show results of continuous data.
0
2
4
6
8
5 15 25 35 45 55 65
Freq
uenc
y
Histogram: Height of Red Trees
Class Midpoints
Histogram ExampleClass
10 but less than 20 15 320 but less than 30 25 630 but less than 40 35 540 but less than 50 45 450 but less than 60 55 2
FrequencyClass Midpoint
Frequency distribution X Tally f Class Boundaries f cf
15 โ 20 / / / / / 5 14.5 โ 20.5 5 521 โ 26 / / / / / / / / / 9 20.5 โ 26.5 9 1427 โ 32 / / / / 4 26.5 โ 32.5 4 1833 โ 38 / / / / / / / / / / 10 32.5 โ 38.5 10 2839 โ 44 / / / / 4 38.5 โ 44.5 4 3245 - 50 / / / / / / / / 8 44.5 โ 50.5 8 40
n = 40
Frequency distribution X Tally f Class Boundaries f cf
15 โ 20 / / / / / 5 14.5 โ 20.5 5 521 โ 26 / / / / / / / / / 9 20.5 โ 26.5 9 1427 โ 32 / / / / 4 26.5 โ 32.5 4 1833 โ 38 / / / / / / / / / / 10 32.5 โ 38.5 10 2839 โ 44 / / / / 4 38.5 โ 44.5 4 3245 - 50 / / / / / / / / 8 44.5 โ 50.5 8 40
n = 40
Class boundaries0
2
4
6
8
10
12
14.5 - 20.5
20.5 - 26.5
26.5 - 32.5
32.5 - 38.5
38.5 - 44.5
44.5 - 50.5Fr
eque
ncy
Frequency polygon- Is constructed by plotting the class
marks against the class frequencies.
5 10 15 20 250123456789
10
Scores
Midpoints
Freq
uenc
y
Frequency distribution X Tally f cf
15 โ 20 / / / / / 5 17.5 521 โ 26 / / / / / / / / / 9 23.5 1427 โ 32 / / / / 4 29.5 1833 โ 38 / / / / / / / / / / 10 35.5 2839 โ 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40
n = 40
Frequency distribution X Tally f cf
15 โ 20 / / / / / 5 17.5 521 โ 26 / / / / / / / / / 9 23.5 1427 โ 32 / / / / 4 29.5 1833 โ 38 / / / / / / / / / / 10 35.5 2839 โ 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40
n = 40
17.5 23.5 29.5 35.5 41.5 47.50
2
4
6
8
10
12
Scores
Midpoints
Freq
uenc
y
Pie graphThis displays data in an easy-to-read โpie-sliceโ format with varying slice sizes.
48%
19%
9%
9%10%
5%
Monthly BudgetRent Food UtilitiesFun Clothes Phone
Frequency distribution X Tally f cf
15 โ 20 / / / / / 5 17.5 521 โ 26 / / / / / / / / / 9 23.5 1427 โ 32 / / / / 4 29.5 1833 โ 38 / / / / / / / / / / 10 35.5 2839 โ 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40
n = 40
Frequency distribution X Tally f cf
15 โ 20 / / / / / 5 17.5 521 โ 26 / / / / / / / / / 9 23.5 1427 โ 32 / / / / 4 29.5 1833 โ 38 / / / / / / / / / / 10 35.5 2839 โ 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40
n = 40
13%
23%
10%25%
10%
20%
Scores
15 - 2021 - 2627 - 3233 - 3839 - 4445 - 50
A PIE graph
Measures of Central
Tendency
Measures of Central Tendency Mean
Median
Mode
Mean Equation :
Where: the sum of the individual values.N- total number of values.
Characteristics of a Mean1. All values are used.2. Most sensitive measures of central
tendency to use for ratio data.3. Influenced by extreme scores.4. The sum of the deviations from the mean is 0.
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
How to Compute for a Population Mean?
Equation:
- the population mean.N- the number of values in the population.x- any particular value.- the sum of the x values in the population.
Find the Population Mean:
Faculty members of (5) colleges:1625204032
Solution:
26.6 27
How to compute for a sample mean?
Equation:
Where:x- the sum of all data valuesn- the number of data items in sample
Find the sample mean
The following are the ages of samples of 8 children in a city:
9 8 1 3 4 5 6 7
Solution:x x xx 5.4x 5
__
__
__
__
__
Mean for ungrouped dataEquation::
Where,:x- sum of the individual value. n- total number of values.
nx
x
:
Example:
Raw scores of 7 students in a 60 item science examination.
Scores: 56, 45, 40, 34, 34, 32, 25
Solution:
x x x x 38
__
__
__
__
Mean for Grouped DataEquation:
x = where:m- the midpoint of the interval,f- the frequency for the interval, โ sum of the product of the midpoint
and the frequency, n- the number of values.
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 1927 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Solution:Class
Intervalf xm fxm
15 โ20 5 17.5 87.521 โ26 9 23.5 211.527 โ32 4 29.5 11833 โ38 10 35.5 35539 โ44 4 41.5 16645 - 50 8 47.5 380
n=40 f(xm)=1318
Solution:Class
Intervalf f
12 โ17 2 14.5 2918 โ23 7 20.5 143.524 โ29 7 26.5 185.530โ35 7 32.5 227.536โ41 7 38.5 269.5 42-47 7 44.5 311.548-53 3 50.5 1151.5
n=40 f()=1366
Solution:
x = x x = 34.15
__
__
__
Frequency Distribution
MEAN
WEIGHTED MEAN COMBINE MEAN
What is the weighted mean ?
WEIGHTED MEAN is a measurement of central
tendency. It represents the average of a given data.
is similar to arithmetic mean or sample mean.
Weighted Mean Formula
ร=โ ๐ ๐ร๐
โ ๐ ๐
1. Multiply
the numbers in your data set by the
weights.
2. Add the
numbers in Step 1 up. Set
this number aside for
a moment.
3. Add up all the
weights
4.Divide the
numbers you
found in Step 2 by the
number you
found in Step 3
Weighted Mean Problems
The marks obtain by 20 students in a Mathematics test are 12,15,18, 10, 20, and the corresponding frequencies are 3,6,5,4,2. Compute the average marks obtained by the students.
Solution : xi fi xi fi
12
15
1810 20
Solution : xi fi xi fi
12 3
15 6
18 5
10 4
20 2
โfi = 20
Solution : xi fi fixi
12 3 3615 6 9018 5 9010 4 4020 2 40
โfi = 20 โ fixi = 296
Substitute the given values
Answer : 14.8
ร=โ ๐ ๐ร๐
โ ๐ ๐
Weighted Mean Problems The number of students absent in a on
few subsequent days are as follows: 1,3,4,4,1,7,5,2,4,3,7,3,4,5,3,5,7,7.
Weighted Mean Problems
Find the average of absent students using the weighted mean formula.
Solution : x i f i
1 22 1
3
4
4 45 37 4
Solution : xi fi fixi
1 2 22 1 23 4 124 4 165 3 157 4 28
โfi = 18 โfixi = 75
Weighted Average with Percentage A student is enrolled in a biology course where the final grade is determined based on the following categories: tests 40%, final exam 25%, quizzes 25%, and homework 10%.
The student has earned the following scores for each category: tests-83, final exam-75, quizzes-90, homework-100.
We need to calculate the student's
overall grade.
COMBINED
MEAN
Three sections of a Statistics class containing 35, 40, and 45 students averaged 80, 85 and 69 respectively on the same final examination.
What is the combined mean for all the three sections .
Number of Students (n)
Average Grade ()
n
35 80 2,800 40 85 3,40045 89 4,005
= 120 = 10,205
MEDIAN
Median For the Ungrouped Data = +
2 ,if N is even
= + N
2 , if N is odd
_________________
_________________
A B40 4034 3434 3445 4556 5632 3225 25
60
A B25= 25= 32= 32= 34= 34= 34= 45= 40= 56= 45= 32= 56= 25=
60=
For Group A, N =7 (odd)
= 2
= 2
= 2
= =12
For Group B, N= 8 (even)
= + 2 = + 2 = + 2 =+ 2
= + 5 2 = == 34.5~ร
Median for the Grouped Data
Class Interval
f Cum f<
15-20 4 421-26 9 1327-32 3 1633-38 10 2639-44 4 3045-50 10 40
N=40
L + - F fm(i)=
Where, L = exact lower limit of the interval
containing the median class F = the sum of all frequencies below L. fm= frequency of interval containing
the median class. N= total number of cases i = class interval
~ร
= 32.5 + - 16 10 = 32.5 + 20 โ 16 10 = 32.5 + 4 10
= 32.5 + 24 10
= 32.5 + 2.4 = 34.9
L = 32.5 F = 16Fm = 10 N = 40 I = 6
_______
_______
__
(6)
____
(6)
(6)
MODEMost Often
What is Mode?
The value that appears most in a given data.
French expression โa la modeโ meaning fashionable.
Types of Mode
Unimodal
Bimodal
Trimodal or Multimodal
Example:Scores of Section A
Scores of Section B
Scores of Section C
25 25 2524 24 2524 24 2520 20 2220 18 2120 18 2116 17 2112 10 1810 9 187 6 18
Merits of Mode
Most typical value of a distribution.
Can be used to describe qualitative distribution.
Demerits of Mode Value of mode cannot always
be determined.
Value of mode is not based on each and every item of the series.
It does not always exist.
How to find the Mode?
Mode for Ungrouped Data
Example:The number of points scored in a series of football game listed below.
7, 13, 18, 24, 9, 3, 18
Solution:Order the scores from least to greatest.
3, 7, 9, 13, 18, 18, 24
Answer:18 - unimodal
Exercises:In a crash test, 11 cases were tested to determine the impact speed required to obtain minimal bumper damage.
24, 15, 18, 20, 18, 22, 24, 26, 18, 26, 24
A marathon race was completed by 5 marathon participants.
2.7, 8.3, 3.5, 5.1, 4.9
On a cold winter day in January, the temperature for 9 North American cities is recorded in Fahrenheit.
-8, 0, -3, 4, 12, 0, 5, -1, 0
The following is the number of problems that Ms. Matty assigned for homework on 10 different days.
8, 11, 9, 14, 11, 9, 15, 9, 18, 11
The number points scored in a series of basketball games is listed below.
8, 19, 14, 19, 14, 24, 8
Mode for Grouped Data
Raw scores of 40 students in a 50-item mathematics quiz.
Computing the RangeThe range is the difference between the highest
score and the lowest score.Range = HS โ LS
R= 50 โ 15 R= 35
Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 + 3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6
Is the groupings defined by the lower and upper limits.Example: LL โ UL
15 โ 20 21 โ 26
27 โ 32
STEPS:
1. Determine the modal class.2. Get the value of .3. Get the value of .4. Get the lower boundary of the
modal class.5. Apply the formula.
FORMULA:
c.i.
WHERE,
โข ๐ - Mode
โข - lower boundary of modal class
โข Modal Class (MC) - category containing highest frequency
โข - difference between frequency of modal class and frequency above it
โข - difference between frequency of modal class and frequency below it
โข c.i. - size of class interval
.
Solution:MC: 33-3810-4 = 6: 10-4 = 6: 33-0.5 = 32.5c.i.: 6Formula:
c.i.
Substitution:= 32.5 + 6 = 32.5 + 6 = 32.5 + 6 = 32.5 + 3= 35.5 or 36
Steps:
1. Determine the modal class.2. Get the value of .3. Get the value of .4. Get the lower boundary of
the modal class.5. Apply the formula.
Formula:
c.i.
Where,
โข - Mode
โข - lower boundary of modal class
โข Modal Class (MC) - category containing highest frequency
โข - difference between frequency of modal class and frequency above it
โข - difference between frequency of modal class and frequency below it
โข c.i. - size of class interval
Exercise:โข Measures of 40 mango leaves in cm.
x f10-14 515-19 220-24 325-29 530-34 235-39 940-44 645-49 350-54 3
n= 40
MEASURES OF POSITION
โข QUARTILE
โข DECILE
โข PERCENTILE
QUARTILE
4
4
4
4
These are values that divide the set of data into four equal parts.
QUARTILE PERCENTAGE
25%
50%
75%
FORMULA: Ungrouped Data
Qk =is the indicated quartileq = 1, 2, 3n = number of cases
Find the quartile two of the following set of scores; 45, 40, 56, 32, 34, 25, and 34
25, 32, 34, 34, 40, 45, 56
(arrange the scores from highest to lowest)
1st 3rd 4th2nd 5th 6th 7th
nth score = = = 4th score = 34
25, 32, 34, 34, 40, 45, 56
/ 4th score
FORMULA: Grouped Data
Qk = the indicated quartilek = 1, 2, and 3Lb = lower boundary of the quartile classcfp= cumulative frequencyfq = frequency of the indicated decile classci = size of the class interval
Raw scores of 40 students in a 50 item math quiz
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
SCORES (X) TALLY FREQUENCY CF <
12- 17 II 2 218- 23 IIII- II 7 924- 29 IIII- II 7 1630- 35 IIII- II 7 2336- 41 IIII -II 7 3042- 47 IIII - II 7 3748- 53 III 3 40
n = 40
= 29.5 + 6 Given: LB = 29.5 = 29.5 + 6 = 20 = 29.5 + 3.4 cf = 16 = 32.9 f = 7 ci = 6
SCORES (X) TALLY FREQUENCY CF <
12- 17 II 2 218- 23 IIII- II 7 924- 29 IIII- II 7 1630- 35 IIII- II 7 2336- 41 IIII -II 7 3042- 47 IIII - II 7 3748- 53 III 3 40
n = 40
๐2
DECILE
10
1010
10
These are values that divide a set of observations into 10 equal parts. Usually denoted by D1,D2,D3,โฆ D9.
DECILE PERCENTAGE 10%
20%30%40%50%60%70%80%90%
FORMULA: Ungrouped Data
is the indicated decilek = 1, 2, 3, 4, 5, 6, 7, 8, 9n = number of cases
25, 32, 34, 34, 40, 45, 46
Raw scores of 7 students 60- item science examination
= = = = 4th score
= 34
25, 32, 34, 34, 40, 45, 56
/ 4th score
FORMULA: Grouped Data
= indicated decile k = 1, 2,3,โฆ9 =lower boundary of the indicated decile classcfp= cumulative frequency fd = frequency of the indicated decile class ci = size of the class interval
Raw scores of 40 students in a 50 item math quiz
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
SCORES (X) TALLY FREQUENCY CF <
12- 17 II 2 218- 23 IIII- II 7 924- 29 IIII- II 7 1630- 35 IIII- II 7 2336- 41 IIII -II 7 3042- 47 IIII - II 7 3748- 53 III 3 40
n = 40
= + Given: = 29.5 + 6 LB = 29.5 = 29.5 + 6 = 20 = 29.5 + 3.4 cf = 16 32.9 f = 7 ci = 6
SCORES (X) TALLY FREQUENCY CF <
12- 17 II 2 218- 23 IIII- II 7 924- 29 IIII- II 7 1630- 35 IIII- II 7 2336- 41 IIII -II 7 3042- 47 IIII - II 7 3748- 53 III 3 40
n = 40
๐ท5
PERCENTILE
100
100100
100
These are values that divide the set of data into 100 equal parts. Usually denoted by ,,,โฆ
PERCENTILE PERCENTAGE
1%
2%
3%
99%
FORMULA: Ungrouped Data
=is the indicated percentile
k = 1, 2, 3,โฆ, 99n = number of cases
Raw scores of 7 students in a 60- item science examination
= = = = 4th score = 34
FORMULA: Grouped Data
= indicated percentileK = 1, 2,3,โฆ99 =lower boundary of the
indicated decile classcfp= cumulative frequencyfd = frequency of the indicated decile classci = size of the class interval
Raw scores of 40 students in a 50 item math quiz
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
SCORES (X) TALLY FREQUENCY CF <
12- 17 II 2 218- 23 IIII- II 7 924- 29 IIII- II 7 1630- 35 IIII- II 7 2336- 41 IIII -II 7 3042- 47 IIII - II 7 3748- 53 III 3 40
n = 40
= ci Given: = 29.5 + 6 LB = 29.5 = 29.5 + 6 = 20 = 29.5 + 3.4 cf = 16= 32.9 f = 7 ci = 6
SCORES (X) TALLY FREQUENCY CF <
12- 17 II 2 218- 23 IIII- II 7 924- 29 IIII- II 7 1630- 35 IIII- II 7 2336- 41 IIII -II 7 3042- 47 IIII - II 7 3748- 53 III 3 40
n = 40
๐50
GRACIAS! HOPE YOUโVE LEARNED.
LOCATION
MOTION
PHYSICAL APPEARANCE
SKIN REACTION TO DIFFERENT CHEMICALS
HEIGHT
WEIGHT
HAIR COLOR
EYE COLOR
IDEAS
VALUES IN LIFE
HEIGHT OF FILIPINO MALE ADULTS 5โ6
=
=
Variation
Mean
No Variation in Cash Flow
Variation in Cash Flow
homogeneous
heterogeneous
No Variability
Variability
RANGE
RANGEa. Range for Ungrouped Data
R= HS-LS
Where,R- range valueHS- highest scoreLS- lowest score
Example:Find the range of the two
groups of score distribution. GROUP A GROUP B
25 2032 3034 3434 3540 4345 4656 60
Example:Find the range of the two
groups of score distribution. GROUP A GROUP B
25 2032 3034 3434 3540 4345 46
56 HS 60 HS
Example:Find the range of the two
groups of score distribution. GROUP A GROUP B
25-LS 20-LS32 3034 3434 3540 4345 46
56-HS 60-HS
RANGERange for Ungrouped Data
For Group A
RA = HS-LS = 56-25RA = 31
RANGERange for Ungrouped Data
For Group BRB = HS-LS
= 60-20RB = 40
RANGEb. Range for Grouped Data
R= HSUB-LSLB
Where,R - range valueHSUB - upper boundary of
highest scoreLSLB - lower boundary of lowest score
Example:Find the value of range
of the scores of 40 students in Science examination test. . x f
15-20 421-26 927-32 335-38 1039-44 445-50 10
n= 40
Range for Grouped Data
R= HSUB-LSLB
= 50.5-14.5R= 36
RANGE
MEAN DEVIATI
ON
MEAN DEVIATIONa. Mean Deviation for Ungrouped
Data
MD=
Where,MD- mean deviation value x- individual score x- sample mean n- number of scores
_
Example:Find the mean deviation of
the scores of 7 students in a Science test.
The scores are 40, 34, 34, 45, 56, 32 and 25
x x-x
40343445563225
|รโร|__
x= n
x= 40+34+34+45+56+32+25 n = 266 7x= 38
________
_
_
_
x x-x
40343445563225
|รโร|__
x x-x
40 234 -434 -445 756 1832 -625 -13
x= 266
|รโร|__
x x-x
40 2 234 -4 434 -4 445 7 756 18 1832 -6 625 -13 13
x= 26654
|รโร|
__
__
MD= n
= 54 7
MD= 7.71
โ_________
MEAN DEVIATION
b. Mean Deviation for Grouped Data
MD= n
_________โ ๐ |ร๐โร|
Where,
MD- mean deviation value f- class frequency
xm- class mark or midpoint of each category
x- mean value n- number of cases
_
f f f15-20 521-26 927-32 433-38 1039-44 4
45-50 8n=40
Example:Find the mean deviation of the given scores below.
f f f15-20 5 17.5
21-26 9 23.5
27-32 4 29.5
33-38 10 35.5
39-44 4 41.5
45-50 8 47.5
n=40
f f f15-20 5 17.5 87.5
21-26 9 23.5 211.5
27-32 4 29.5 118
33-38 10 35.5 355
39-44 4 41.5 166
45-50 8 47.5 380
n=40
n
= 1318 40
= 32.95
f f f
15-20 5 17.5 87.5
21-26 9 23.5 211.5
27-32 4 29.5 118
33-38 10 35.5 355
39-44 4 41.5 166
45-50 8 47.5 380
n=40
f f f15-20 5 17.5 87.5 -15.45
21-26 9 23.5 211.5 -9.45
27-32 4 29.5 118 -3.45
33-38 10 35.5 355 2.55
39-44 4 41.5 166 8.55
45-50 8 47.5 380 14.55
n=40
f f f15-20 5 17.5 87.5 -15.45 15.45
21-26 9 23.5 211.5 -9.45 9.45
27-32 4 29.5 118 -3.45 3.45
33-38 10 35.5 355 2.55 2.55
39-44 4 41.5 166 8.55 8.55
45-50 8 47.5 380 14.55 14.55
n=40
f f f15-20 5 17.5 87.5 -15.45 15.45 77.25
21-26 9 23.5 211.5 -9.45 9.45 85.05
27-32 4 29.5 118 -3.45 3.45 13.8
33-38 10 35.5 355 2.55 2.55 25.5
39-44 4 41.5 166 8.55 8.55 34.2
45-50 8 47.5 380 14.55 14.55 116.4
n=40
MD= n
= 352.2 40
MD= 8.805
VARIANCE & STANDARD DEVIATION
VARIANCEa. Variance of Ungrouped Data
Population variance = Sample variance
=
Example:Using the data, find the variance and standard deviation of the scores of 7 students in a science quiz.
x40343445563225
x
40 234 -434 -445 756 1832 -625 -13
= 38
x
40 2 4
34 -4 16
34 -4 16
45 7 49
56 18 324
32 -6 36
25 -13 169
= 38
Population variance
=
= 614 7
=87.71
Sample variance
= = 614
7-1 = 614
6 = 102.33
VARIANCEb.Variance of Grouped Data
Population variance = Sample variance
=
Score distribution of the test results of 40 students in a science class consisting of 50 items. Solve the variance and standard deviation. f f - f
15-20 5
21-26 9
27-32 4
33-38 10
39-44 4
45-50 8
N=40
f f - f
15-20 5 17.5
21-26 9 23.5
27-32 4 29.5
33-38 10 35.5
39-44 4 41.5
45-50 8 47.5
N=40
f f - f15-20 5 17.5 87.521-26 9 23.5 211.527-32 4 29.5 11833-38 10 35.5 35539-44 4 41.5 16645-50 8 47.5 380
N=40
f f - f
15-20 5 17.5 87.5 32.95
21-26 9 23.5 211.5 32.95
27-32 4 29.5 118 32.95
33-38 10 35.5 355 32.95
39-44 4 41.5 166 32.95
45-50 8 47.5 380 32.95
N=40
f f - f
15-20 5 17.5 87.5 32.95 -15.45
21-26 9 23.5 211.5 32.95 -9.45
27-32 4 29.5 118 32.95 -3.45
33-38 10 35.5 355 32.95 2.55
39-44 4 41.5 166 32.95 8.55
45-50 8 47.5 380 32.95 14.55
N=40
f f - f
15-20 5 17.5 87.5 32.95 -15.45 238.70
21-26 9 23.5 211.5 32.95 -9.45 89.30
27-32 4 29.5 118 32.95 -3.45 11.90
33-38 10 35.5 355 32.95 2.55 6.50
39-44 4 41.5 166 32.95 8.55 73.10
45-50 8 47.5 380 32.95 14.55 211.70
N=40
f f - f
15-20 5 17.5 87.5 32.95 -15.45 238.70 1193.5
21-26 9 23.5 211.5 32.95 -9.45 89.30 803.7
27-32 4 29.5 118 32.95 -3.45 11.90 97.6
33-38 10 35.5 355 32.95 2.55 6.50 65
39-44 4 41.5 166 32.95 8.55 73.10 292.4
45-50 8 47.5 380 32.95 14.55 211.70 1693.6
N=40= 4145.8
Population variance
=
= 4145.8 40
=103.645
Sample variance= = 4145.8
40-1 = 4145.8
39 = 106.30
STANDARD DEVIATIONa. Standard Deviation of Ungrouped Data Population Standard Deviation = Sample Standard Deviation s=
Population Standard Deviation =
=
=
= 9.37
Sample Standard Deviation s=
=
=
s=10.12
STANDARD DEVIATIONb. Standard Deviation of
Grouped Data Population Standard Deviation = Sample Standard Deviation s=
Population Standard Deviation =
=
=
=10.18
Sample Standard Deviation s=
=
=
s=10.31
Section A Section B Section C
12 12 12
12 12 12
14 12 12
15 13 12
17 13 12
18 14 12
18 17 13
18 20 26
19 20 26
23 28 26
23 28 26
30 30 30
X = 18.25 X = 18.25 X = 18.25
S= 5.15 S= 6.92 S= 7.63
COEFFICIENT OF
VARIATION
GROUP 1: = 156cm , s= 6
GROUP 2: = 156cm , s=10more varied
BOYS : = 160 lbs s= 8
GIRLS : = 100 lbs s= 9
COEFFICIENT OF VARIATIONb. Formula for population
Where,s- standard variation- mean
COEFFICIENT OF VARIATIONa. Formula for population
GROUP s CV
A 38 9.37
B 38.29 11.86
COEFFICIENT OF VARIATION๐ถ๐=ยฟ
๐๐ ร 100%
Group A
Group B
COEFFICIENT OF VARIATIONa. Formula for population
B 38.29 11.86 31
COEFFICIENT OF VARIATIONb. Formula for sample
Where,s- standard variation- mean
COEFFICIENT OF VARIATIONa. Formula for sample
10.12B 38.29 12.82
COEFFICIENT OF VARIATION๐ถ๐=ยฟ
๐ร ร 100%
Group A
Group B
COEFFICIENT OF VARIATIONa. Formula for sample
10.12 27B 38.29 12.82 33
QUARTILE
DEVIATION
QUARTILE DEVIATION Formula:
Where, - quartile deviation - 3rd quartile - 1st quartile
Inter- quartile Range
QUARTILE DEVIATION For ungrouped data
Find the quartile deviation of the scores of 7 students in a Science test.
The scores are 40, 34, 34, 45, 56, 32 and 25
QUARTILE DEVIATION Scores: 25 32 34 34 40 45 56
QUARTILE DEVIATION Scores: 25 32 34 34 40 45 56
๐3=3๐
4h๐ก
item
Thus, = 45
QUARTILE DEVIATION Scores: 25 32 34 34 40 45 56
๐3=3๐
4h๐ก ๐
1=๐4
h๐ก
item
Thus, = 45
item
Thus, = 32
QUARTILE DEVIATION Scores: 25 32 34 34 40 45 56
= 45= 32
Formula:
Class interval
f cfClass boundariesLower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
Find the quartile deviation of the given scores below.
QUARTILE DEVIATION
๐3=3๐
4h๐ก
item
Formula :
Class interval f cf
Class boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
Formula :
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=30
Hence,
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=30
Hence,๐๐ 3=ยฟ ยฟ
cf๐=ยฟ ยฟ
๐ ๐3=ยฟยฟ
i=
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=30
Hence,๐๐ 3=35.5
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=30
Hence,๐๐ 3=35.5
cf๐=23
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=30
Hence,๐๐ 3=35.5
cf๐=23
๐ ๐3=7
i= 6
Formula :
QUARTILE DEVIATION
๐1=๐
4h๐ก
item
Formula :
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=10
Hence,๐๐ 3=ยฟ ยฟ
cf๐=ยฟ ยฟ
๐ ๐3=ยฟยฟ
i=
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐1=10
Hence,
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐1=10
Hence,๐๐ 1=23.5
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐1=10
Hence,๐๐ 1=23.5
cf๐=9
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
๐1=10
Hence,๐๐ 1=23.5
cf๐=9
๐ ๐1=7
i= 6
Formula :
QUARTILE DEVIATION
= 41.5= 24.34
Formula:
PERCENTILE
RANGE
PERCENTILE RANGEFormula :
PR=
Where, PR - percentile range -90th percentile - 10th percentile
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 1927 35 45 48 20 38 39 1844 22 46 26 36 29 15 2150 47 34 26 37 25 33 4922 33 44 38 46 41 37 32
Raw scores of 40 students in a 50-item science quiz.
15 17 18 19 20 21 22 2223 25 25 25 26 26 27 2930 32 33 33 33 34 35 3637 37 38 38 39 41 44 4445 45 46 46 47 48 49 50
PERCENTILE RANGE
item , which is 46
๐ท๐๐=๐๐๐ต
๐๐๐ ๐๐
15 17 18 19 20 21 22 2223 25 25 25 26 26 27 2930 32 33 33 33 34 35 3637 37 38 38 39 41 44 4445 45 46 46 47 48 49 50
36th item
PERCENTILE RANGE
item , which is 46
๐ท๐๐=๐๐๐ต
๐๐๐ ๐๐๐ท
๐๐=๐๐๐ต๐๐๐ ๐๐
item , which is 19
15 17 18 19 20 21 22 2223 25 25 25 26 26 27 2930 32 33 33 33 34 35 3637 37 38 38 39 41 44 4445 45 46 46 47 48 49 50
36th item
4th item
PERCENTILE RANGE
Formula:
PR=
= 46-19PR=27
COEFFICIENT
VARIATION
COEFFICIENT VARIATIONb. Formula for population
Where,s- standard variation- mean
COEFFICIENT VARIATIONa. Formula for population
GROUP s CV
A 38 9.37
B 38.29 11.86
COEFFICIENT VARIATION๐ถ๐=ยฟ
๐๐ ร 100%
Group A
Group B
COEFFICIENT VARIATIONa. Formula for population
B 38.29 11.86 31
COEFFICIENT VARIATIONb. Formula for sample
Where,s- standard variation- mean
COEFFICIENT VARIATIONa. Formula for sample
10.12B 38.29 12.82
COEFFICIENT VARIATION๐ถ๐=ยฟ
๐ร ร 100%
Group A
Group B
COEFFICIENT VARIATIONa. Formula for sample
10.12 27B 38.29 12.82 33
QUARTILE
DEVIATION
QUARTILE DEVIATION Formula:
Where, - quartile deviation - 3rd quartile - 1st quartile
QUARTILE DEVIATION For ungrouped data
Find the quartile deviation of the scores of 7 students in a Science test.
The scores are 40, 34, 34, 45, 56, 32 and 25
QUARTILE DEVIATION Scores: 25 32 34 34 40 45 56
QUARTILE DEVIATION Scores: 25 32 34 34 40 45 56
๐3=3๐
4h๐ก
item
Thus, = 45
QUARTILE DEVIATION Scores: 25 32 34 34 40 45 56
๐3=3๐
4h๐ก ๐
1=๐4
h๐ก
item
Thus, = 45
item
Thus, = 32
QUARTILE DEVIATION Scores: 25 32 34 34 40 45 56
= 45= 32
Formula:
Class interval
f cfClass boundariesLower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
Find the quartile deviation of the given scores below.
QUARTILE DEVIATION
๐3=3๐
4h๐ก
item
Formula :
Class interval f cf
Class boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
Formula :
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=30
Hence,
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=30
Hence,๐๐ 3=35.5
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=30
Hence,๐๐ 3=35.5
cf๐=23
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐3=30
Hence,๐๐ 3=35.5
cf๐=23
๐ ๐3=7
i= 6
Formula :
QUARTILE DEVIATION
๐1=๐
4h๐ก
item
Formula :
Class interval f cf
Class boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
Formula :
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐1=10
Hence,
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐1=10
Hence,๐๐ 1=23.5
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐1=10
Hence,๐๐ 1=23.5
cf๐=9
Class interval
f cfClass
boundariesLower Upper
12-17 2 14.5 2 11.5 17.518-23 7 20.5 9 17.5 23.524-29 7 26.5 16 23.5 29.530-35 7 32.5 23 29.5 35.536-41 7 38.5 30 35.5 41.542-47 7 44.5 37 41.5 47.548-53 3 50.5 40 47.5 53.5
N=40
๐1=10
Hence,๐๐ 1=23.5
cf๐=9
๐ ๐1=7
i= 6
Formula :
QUARTILE DEVIATION
= 41.5= 24.34
Formula:
PERCENTILE
RANGE
PERCENTILE RANGEFormula :
PR=
Where, PR - percentile range -90th percentile - 10th percentile
Raw scores of 40 students in a 50-item mathematics quiz.
17 25 30 33 25 45 23 1927 35 45 48 20 38 39 1844 22 46 26 36 29 15 2150 47 34 26 37 25 33 4922 33 44 38 46 41 37 32
Raw scores of 40 students in a 50-item science quiz.
15 17 18 19 20 21 22 2223 25 25 25 26 26 27 2930 32 33 33 33 34 35 3637 37 38 38 39 41 44 4445 45 46 46 47 48 49 50
PERCENTILE RANGE
item , which is 46
๐ท๐๐=๐๐๐ต
๐๐๐ ๐๐
15 17 18 19 20 21 22 2223 25 25 25 26 26 27 2930 32 33 33 33 34 35 3637 37 38 38 39 41 44 4445 45 46 46 47 48 49 50
36th item
PERCENTILE RANGE
item , which is 46
๐ท๐๐=๐๐๐ต
๐๐๐ ๐๐๐ท
๐๐=๐๐๐ต๐๐๐ ๐๐
item , which is 19
15 17 18 19 20 21 22 2223 25 25 25 26 26 27 2930 32 33 33 33 34 35 3637 37 38 38 39 41 44 4445 45 46 46 47 48 49 50
36th item
4th item
PERCENTILE RANGE
Formula:
PR=
= 46-19PR=27