Assessment of Learning 2 (Overview)

What is Frequency?• 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 Drivers’ Ages

Parts of a frequency distribution :• Class limit• Class size• Class boundaries• Class marks

Class limit- Is the groupings defined by

the lower and upper limits.Example: LL – UL

10 – 14 15 – 19

20 – 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 – UL

10 – 14 15 – 19

20 – 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 – UL 10 – 14 9.5 – 14.5 15 – 19 14.5 – 19.5

20 – 24 19.5 – 24.5

Class marks- Are the midpoint of the lower

and upper class limits.- Formula is =

Example: LL – UL 10 – 14 -

15 – 19 - 20 – 24 -

Class marks- Are the midpoint of the lower

and upper class limits.- Formula is =

Example: LL – UL 10 – 14 12

15 – 19 17 20 – 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


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


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


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


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 – LS R= 50 – 15 R= -

Computing the RangeThe range is the difference

between the highest score and the lowest score.

Range = HS – LS R= 50 – 15 R= 35

Compute the

Range;R=HS-LS


Set up the class

limits.

Make a tally.

Count the

frequency (f).

Compute the

Range;R=HS-LS


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


Solve the value of k:k = 1 + 3.3 log nk = 1 + 3.3 log 40k = 1 +

3.3(1.60205991)



3.3(1.60205991)k = 1 + 5.286797971



3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971



3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6



3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6


Find the class size:

c.i =


3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6



c.i = c.i =


3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6



c.i = c.i = c.i = 5.833


3.3(1.60205991)k = 1 + 5.286797971k = 6.286797971k = 6



c.i = c.i = c.i = 5.833c.i = 6

From Assessment of Learning 1, Yonardo Gabuyo

Solve the value of k:k =


Solve the value of k:k = k =


Solve the value of k:k = k = k = 6.32455532


Solve the value of k:k = k = k = 6.32455532k = 6



c.i = c.i = c.i = 5.833c.i = 6

Compute the

Range;R=HS-LS


Set up the class

limits.

Make a tally.

Count the

frequency (f).

Compute the

Range;R=HS-LS


Set up the class

limits.

Make a tally.

Count the

frequency (f).

Set up the class limits.X

15 – 2021 – 2627 – 3233 – 3839 – 4445 - 50

Compute the

Range;R=HS-LS


Set up the class

limits.

Make a tally.

Count the

frequency (f).

Compute the

Range;R=HS-LS


Set up the class

limits.

Make a tally.

Count the

frequency (f).

Make a tally X Tally

15 – 20 / / / / /21 – 26 / / / / / / / / /27 – 32 / / / /33 – 38 / / / / / / / / / /39 – 44 / / / /45 - 50 / / / / / / / /


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


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


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


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


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


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


Set up the class

limits.

Make a tally.

Count the

frequency (f).

Compute the

Range;R=HS-LS


Set up the class

limits.

Make a tally.

Count the

frequency (f).

Indicate the frequency. X Tally Frequency (f)

15 – 20 / / / / / 521 – 26 / / / / / / / / / 927 – 32 / / / / 433 – 38 / / / / / / / / / / 1039 – 44 / / / / 445 - 50 / / / / / / / / 8

n = 40

Frequency distribution X Tally f cf

15 – 20 / / / / / 5 17.5 521 – 26 / / / / / / / / / 9 23.5 1427 – 32 / / / / 4 29.5 1833 – 38 / / / / / / / / / / 10 35.5 2839 – 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:

• Bar graph• Histogram• Frequency polygon• 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


15 – 20 / / / / / 5 17.5 521 – 26 / / / / / / / / / 9 23.5 1427 – 32 / / / / 4 29.5 1833 – 38 / / / / / / / / / / 10 35.5 2839 – 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40

n = 40


15 – 20 / / / / / 5 17.5 521 – 26 / / / / / / / / / 9 23.5 1427 – 32 / / / / 4 29.5 1833 – 38 / / / / / / / / / / 10 35.5 2839 – 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 – 20 / / / / / 5 14.5 – 20.5 5 521 – 26 / / / / / / / / / 9 20.5 – 26.5 9 1427 – 32 / / / / 4 26.5 – 32.5 4 1833 – 38 / / / / / / / / / / 10 32.5 – 38.5 10 2839 – 44 / / / / 4 38.5 – 44.5 4 3245 - 50 / / / / / / / / 8 44.5 – 50.5 8 40

n = 40

Frequency distribution X Tally f Class Boundaries f cf

15 – 20 / / / / / 5 14.5 – 20.5 5 521 – 26 / / / / / / / / / 9 20.5 – 26.5 9 1427 – 32 / / / / 4 26.5 – 32.5 4 1833 – 38 / / / / / / / / / / 10 32.5 – 38.5 10 2839 – 44 / / / / 4 38.5 – 44.5 4 3245 - 50 / / / / / / / / 8 44.5 – 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


15 – 20 / / / / / 5 17.5 521 – 26 / / / / / / / / / 9 23.5 1427 – 32 / / / / 4 29.5 1833 – 38 / / / / / / / / / / 10 35.5 2839 – 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40

n = 40


15 – 20 / / / / / 5 17.5 521 – 26 / / / / / / / / / 9 23.5 1427 – 32 / / / / 4 29.5 1833 – 38 / / / / / / / / / / 10 35.5 2839 – 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 ‘pie-slice’ format with varying slice sizes.

48%

19%

9%

9%10%

5%

Monthly BudgetRent Food UtilitiesFun Clothes Phone


15 – 20 / / / / / 5 17.5 521 – 26 / / / / / / / / / 9 23.5 1427 – 32 / / / / 4 29.5 1833 – 38 / / / / / / / / / / 10 35.5 2839 – 44 / / / / 4 41.5 3245 - 50 / / / / / / / / 8 47.5 40

n = 40


15 – 20 / / / / / 5 17.5 521 – 26 / / / / / / / / / 9 23.5 1427 – 32 / / / / 4 29.5 1833 – 38 / / / / / / / / / / 10 35.5 2839 – 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.


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, – sum of the product of the midpoint

and the frequency, n- the number of values.


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 –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(xm)=1318

Solution:Class

Intervalf f

12 –17 2 14.5 2918 –23 7 20.5 143.524 –29 7 26.5 185.530–35 7 32.5 227.536–41 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

×=∑ 𝑓 𝑖×𝑖

∑ 𝑓 𝑖

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

∑fi = 20

Solution : xi fi fixi

12 3 3615 6 9018 5 9010 4 4020 2 40

∑fi = 20 ∑ fixi = 296

Substitute the given values

Answer : 14.8

×=∑ 𝑓 𝑖×𝑖

∑ 𝑓 𝑖

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

∑fi = 18 ∑fixi = 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~×

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

~×

= 32.5 + - 16 10 = 32.5 + 20 – 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 “a la mode” 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


Computing the RangeThe range is the difference between the highest

score and the lowest score.Range = HS – LS

R= 50 – 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 – UL

15 – 20 21 – 26

27 – 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,

• 𝒙 - Mode

• - lower boundary of modal class

• Modal Class (MC) - category containing highest frequency

• - difference between frequency of modal class and frequency above it

• - difference between frequency of modal class and frequency below it

• 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,

• - Mode

• - lower boundary of modal class

• Modal Class (MC) - category containing highest frequency

• - difference between frequency of modal class and frequency above it

• - difference between frequency of modal class and frequency below it

• c.i. - size of class interval

Exercise:• 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

• QUARTILE

• DECILE

• 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



n = 40

𝑄2

DECILE

10

1010

10

These are values that divide a set of observations into 10 equal parts. Usually denoted by D1,D2,D3,… D9.

DECILE PERCENTAGE 10%

20%30%40%50%60%70%80%90%


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


= indicated decile k = 1, 2,3,…9 =lower boundary of the indicated decile classcfp= cumulative frequency fd = frequency of the indicated decile class ci = size of the class interval


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



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



n = 40

𝐷5

PERCENTILE

100

100100

100

These are values that divide the set of data into 100 equal parts. Usually denoted by ,,,…

PERCENTILE PERCENTAGE

1%

2%

3%

99%


=is the indicated percentile

k = 1, 2, 3,…, 99n = number of cases

Raw scores of 7 students in a 60- item science examination

= = = = 4th score = 34


= indicated percentileK = 1, 2,3,…99 =lower boundary of the

indicated decile classcfp= cumulative frequencyfd = frequency of the indicated decile classci = size of the class interval


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



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



n = 40

𝑃50

GRACIAS! HOPE YOU’VE 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 5’6

=

=

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



25 2032 3034 3434 3540 4345 46

56 HS 60 HS



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

|×−×|__

x= n

x= 40+34+34+45+56+32+25 n = 266 7x= 38

________

_

_

_

x x-x

40343445563225

|×−×|__

x x-x

40 234 -434 -445 756 1832 -625 -13

x= 266

|×−×|__

x x-x

40 2 234 -4 434 -4 445 7 756 18 1832 -6 625 -13 13

x= 26654

|×−×|

__

__

MD= n

= 54 7

MD= 7.71

−_________

MEAN DEVIATION

b. Mean Deviation for Grouped Data

MD= n

_________∑ 𝒇 |×𝒎−×|

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 VARIATION𝐶𝑉=¿

𝑆𝜇 × 100%

Group A

Group B

COEFFICIENT OF VARIATIONa. Formula for population

B 38.29 11.86 31

COEFFICIENT OF VARIATIONb. Formula for sample


COEFFICIENT OF VARIATIONa. Formula for sample

10.12B 38.29 12.82

COEFFICIENT OF VARIATION𝐶𝑉=¿

𝑆× × 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.


QUARTILE DEVIATION Scores: 25 32 34 34 40 45 56


𝑄3=3𝑁

4h𝑡

item

Thus, = 45


𝑄3=3𝑁

4h𝑡 𝑄

1=𝑁4

h𝑡

item

Thus, = 45

item

Thus, = 32


= 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

𝑄3=3𝑁

4h𝑡

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

𝑄3=30

Hence,

Class interval

f cfClass


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

𝑄3=30

Hence,𝑙𝑄 3=¿ ¿

cf𝑏=¿ ¿

𝑓 𝑄3=¿¿

i=

Class interval

f cfClass


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

𝑄3=30

Hence,𝑙𝑄 3=35.5

Class interval

f cfClass


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

𝑄3=30


cf𝑏=23

Class interval

f cfClass


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

𝑄3=30


cf𝑏=23

𝑓 𝑄3=7

i= 6

Formula :

QUARTILE DEVIATION

𝑄1=𝑁

4h𝑡

item

Formula :

Class interval

f cfClass


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

𝑄3=10

Hence,𝑙𝑄 3=¿ ¿

cf𝑏=¿ ¿

𝑓 𝑄3=¿¿

i=

Class interval

f cfClass


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

𝑄1=10

Hence,

Class interval

f cfClass


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

𝑄1=10


Class interval

f cfClass


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

𝑄1=10


cf𝑏=9

Class interval

f cfClass


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

𝑄1=10


cf𝑏=9

𝑓 𝑄1=7

i= 6

Formula :

QUARTILE DEVIATION

= 41.5= 24.34

Formula:

PERCENTILE

RANGE

PERCENTILE RANGEFormula :

PR=

Where, PR - percentile range -90th percentile - 10th percentile


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

𝑷𝟗𝟎=𝟗𝟎𝑵

𝟏𝟎𝟎 𝒕𝒉

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

𝑷𝟗𝟎=𝟗𝟎𝑵

𝟏𝟎𝟎 𝒕𝒉𝑷

𝟏𝟎=𝟏𝟎𝑵𝟏𝟎𝟎 𝒕𝒉

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


COEFFICIENT VARIATIONa. Formula for population

GROUP s CV

A 38 9.37

B 38.29 11.86

COEFFICIENT VARIATION𝐶𝑉=¿

𝑆𝜇 × 100%

Group A

Group B

COEFFICIENT VARIATIONa. Formula for population

B 38.29 11.86 31

COEFFICIENT VARIATIONb. Formula for sample


COEFFICIENT VARIATIONa. Formula for sample

10.12B 38.29 12.82

COEFFICIENT VARIATION𝐶𝑉=¿

𝑆× × 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.




𝑄3=3𝑁

4h𝑡

item

Thus, = 45


𝑄3=3𝑁

4h𝑡 𝑄

1=𝑁4

h𝑡

item

Thus, = 45

item

Thus, = 32


= 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

𝑄3=3𝑁

4h𝑡

item

Formula :

Class interval f cf


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


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

𝑄3=30

Hence,

Class interval

f cfClass


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

𝑄3=30


Class interval

f cfClass


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

𝑄3=30


cf𝑏=23

Class interval

f cfClass


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

𝑄3=30


cf𝑏=23

𝑓 𝑄3=7

i= 6

Formula :

QUARTILE DEVIATION

𝑄1=𝑁

4h𝑡

item

Formula :

Class interval f cf


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


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

𝑄1=10

Hence,

Class interval

f cfClass


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

𝑄1=10


Class interval

f cfClass


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

𝑄1=10


cf𝑏=9

Class interval

f cfClass


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

𝑄1=10


cf𝑏=9

𝑓 𝑄1=7

i= 6

Formula :

QUARTILE DEVIATION

= 41.5= 24.34

Formula:

PERCENTILE

RANGE

PERCENTILE RANGEFormula :

PR=

Where, PR - percentile range -90th percentile - 10th percentile


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

𝑷𝟗𝟎=𝟗𝟎𝑵

𝟏𝟎𝟎 𝒕𝒉

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

𝑷𝟗𝟎=𝟗𝟎𝑵

𝟏𝟎𝟎 𝒕𝒉𝑷

𝟏𝟎=𝟏𝟎𝑵𝟏𝟎𝟎 𝒕𝒉

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

Education

Assessment of Learning 2 (Overview)