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Mathematics ss2 third term

Mathematics ss2 third term

Statistics 1

CONTENT:

(a) Meaning and computations of mean, median and mode of ungrouped data.

(b) Determination of the mean, median and the mode of grouped frequency data.

(c) Comparison of mean, mode and median.

(d) Rate and mixtures.

 

Meaning and Computation of Mean of Ungrouped Data

The mean, median and the mode are called measures of central tendency or measures of location. The mean is also known as the average, the median is the middle number while the mode is the most frequent element or data.

THE ARITHMETIC MEAN:

The arithmetic mean is the sum of the ungroup of items divided by the number of it. The mean of an ungrouped data can be calculated by using the formula;

x¯=∑xn, (when n is small)

(where the symbol ∑ is called sigma meaning summation of all the given data)

Also, Mean, x¯=∑fx∑f (when n is large)

∑fx= Sum of the product of scores and their corresponding frequencies

∑f= Sum of the frequencies

Example 1:

Find the arithmetic mean of the numbers 42, 50, 59, 38, 41, 86 and 56

Solution: Add all the numbers and divide by 7

x¯=42+50+59+38+41+86+567=3787∴x¯=54

Example 2:

The table below gives the frequency distribution of marks obtained by some students in a scholarship examination.

	Scores (x)	15	25	35	45	55	65	75
	Frequency	1	4	12	24	18	8	3

Calculate, correct to 3 significant figures the mean mark of the distribution (WAEC SSCE)

Solution:         

	Scores (x)	Frequency (f)	fx
	15	1	15
	25	4	100
	35	12	420
	45	24	1080
	55	18	990
	65	8	520
	75	3	225
		∑fx=70	∑fx=3350

x¯=∑fx∑fx¯=335070x¯=47.857143x¯=47.9 3 s.f.

Method 2:

Mean x¯=∑fx∑f=(1×15)+(4×25)+(12×35)+(24×45)+(18×55)+(8×65)+(3×75)1+4+12+24+18+8+3=15+100+420+1080+990+520+2251+4+12+24+18+8+3=335070=47.8571=47.9 3 s.f.

Example 3:

The table below shows the scores of some students in a quiz

	Scores	1	2	3	4	5	6
	Frequency	1	4	5	x	2	2

If the mean score is 3.5, calculate the value of x.

Solution:                         

	x	f	fx
	1	1	1
	2	4	8
	3	5	15
	4	x	4x
	5	2	10
	6	2	12
		∑f=14+x	∑fx=46+4x

Since, mean; x¯=∑fx∑f

But, x¯=3.5⇒3.5=46+4x14+x

On cross multiplying,

3.5(14+x)=46+4x72(14+x)=46+4x7(14+x)=2(46+4x)98+7x=72+8x98−72=8x−7x6=x∴x=6

Mathematics ss2 third term

Example 4:

The table below shows the mark distribution of an English language test in which the mean mark is 3. Find the value of y.

	Marks (x)	1	2	3	4	5
	Frequency (f)	y	3	y+3	3	4 − y

Solution:

Mean; x¯=∑fx∑f

	x	f	fx
	1	Y	y
	2	3	6
	3	y+3	3y + 9
	4	3	12
	5	4 − y	20 − 5y
		∑f=y+13	∑fx=47−y

But, mean; x¯=3

So we have that,  3=47−y13+y

On cross multiplying,

3(13+y)=47−y39+3y=47−y3y+y=47−394y=8∴y=2

Class Activity:

The table below shows the frequency distribution of marks obtained by a group of students in a test. If the mean is 5, calculate the value of x.

	Marks	3	4	5	6	7	8
	Frequency	5	x –1	x	9	4	1

Meaning and computation of median of ungrouped data

The median is the value of the middle item when the items are arranged in order of magnitude either ascending or descending order.

Example 1;

Find the median of the following set of numbers; 16,13,10,23,36,9,8,48,24

Solution:  Arrange in (either ascending or descending order)

8,9,10,13,[16],23,24,36,48

The middle number is 16

∴ Median =16

Median from frequency distribution (i.e when n is large)

Median =(n+12)th, when n is odd

Median =(n2)th+(n2+1)th2 when n is even

Example 2:

The table below shows the distribution of marked scored by some students in a math test

	Marks %	22	24	36	42	45	48	56	60
	Frequency	11	2	7	13	10	3	9	5

Solution:

To find the median, a cumulative frequency table is needed.

	Marks % (x)	Frequency (f)	Cumulative frequency
	22	11	11
	24	2	13
	36	7	20
	42	13	33
	45	10	43
	48	3	46
	56	9	55
	60	5	60

From the table, there are 60 members as indicated by the cumulative frequency.

Since 60 is even, Median =(n2)th+(n2+1)th2=(602)th+(602+1)th2=30th+31st2

The 30^th member is 42% and the 31^st member is 42%

∴ Median =42%+42%2=842=42%

Example 3:

Calculate the median age from the following data.

	Age (yrs)	10	12	13	14	16	17	18	19
	No. of students	7	15	11	7	12	9	4	6

Solution:

	Ages (yrs)	No. of students	Cumulative frequency
	10	7	7
	12	15	22
	13	11	33
	14	7	40
	16	12	52
	17	9	61
	18	4	65
	19	6	71

Since 71 is odd,

Mathematics ss2 third term

Median =(n+12)th member

=(71+12)th=722th

=36th member

The 36^th member falls within the cumulative frequency of up to 40 and this is under 14 years.

∴ Median =14 years

Class Activity:

Calculate the median of the distribution below.

	Marks (x)	10	20	30	40	50
	Frequency (f)	13	18	34	60	10

Meaning and Computation of Mode of Ungrouped Data

The mode of a given data is the item which occurs most often in the distribution

Example 1:

The record of the marks scored by a number of students in an oral test in economics is as follows;

10, 10, 5, 9, 15, 10, 20, 10, 9, 5, 9, 10, 25, 9, 5, 25. Find the modal mark

Solution:

	Marks	5	9	10	15	20	25
	Frequency	3	4	5	1	1	2

From the table above, the highest frequency is 5 and this corresponds to a mark of 10.

The mode is 10

Example 2:

For a class of 30 students, the scores on a maths test out of 20 marks were as follows:

8106108814410461061441218610161210141416141418144

Solution:

	Marks	Frequency
	4	4
	6	4
	8	3
	10	6
	12	2
	14	7
	16	2
	18	2

The highest frequency is 7; Modal score =14

Class Activity:

1. Find the mode of the following distributions.

	Age (years)	13	14	15	16	17	18
	Frequency	3	10	15	21	5	5

2. The table below shows the distribution of test scores in a class

	Scores (x)	No. of pupils (f)
	1	1
	2	1
	3	5
	4	3
	5	k2+1
	6	0
	7	6
	8	2
	9	3
	10	4

If the mean score of the test is 6, find the (a) values of k  (b) median score

Mean Of Grouped Data

Mean for grouped data can be calculated in two ways;

(i) Mean for problems without assumed mean

x¯=∑fx∑f

Where x is the class mark or class midpoint

(ii) Mean of problems with assumed mean

x¯=A+∑fd∑f, where A= assumed mean; d= deviation from mean (x−A)

Mathematics ss2 third term

Example:

1. The weights to the nearest kilogram of a group of 50 students in a college of technology are given below:

6547735366705356595260724863494658515062516758487155626372585964656753637062566968574961635356646459

(a) Prepare a grouped frequency table with class intervals 45–49, 50–54, 55–59, etc

(b) Without the method of assumed, calculate the mean of the grouped data correct to one decimal place.

(c) Using an assumed mean of 62, calculate the mean of the grouped data, correct to one decimal place. (WAEC SSCE)

Solution:

(a)

	Class interval	Frequency
	45 – 49	6
	50 – 54	9
	55 – 59	10
	60 – 64	12
	65 – 69	7
	70 – 74	6

(b) Mean; x¯=∑fx∑f

	Class interval	Class mark (x)	Frequency (f)	fx
	45 – 49	47	6	282
	50 – 54	52	9	468
	55 – 59	57	10	570
	60 – 64	62	12	744
	65 – 69	67	7	469
	70 – 74	72	6	432
			∑f=50	∑fx=2965

x¯=∑fx∑f=296550=59.3 1 d.p.

(c) Mean; x¯=A+∑fd∑f, where d=(x−A) but, A=62

	Class interval	Class mark (x)	Frequency (f)	d=x−Afd	fd
	45 – 49	47	6	–15	– 90
	50 – 54	52	9	–10	–90
	55 – 59	57	10	–5	–50
	60 – 64	62	12	0	0
	65 – 69	67	7	5	35
	70 – 74	72	6	10	60
			∑f=50		∑fd=−135

x¯=A+∑fd∑f=62+(−13550)=62−2.7=59.3 1 d.p.

Class Activity:

The table below gives the masses in kg of 35 students in a particular school. (NECO SSCE)

4550475158436152474254504851515248404149575344505559616062716566687060

(a) Using the above given data, construct a group frequency table with class interval 40 – 44, 45 – 49, 50 – 54 etc

(b) From the data above, calculate the mean of the distributions

(c) Using assumed mean of 52, calculate correct to two decimal places the mean of the distribution

Mathematics ss2 third term

The Median of a Grouped Data

The median formula for grouped data is given as;

Median =L1+[n2−Cfbfm]C

Where; L1= lower class boundary of the median class

n= total frequency

Cfb= cumulative frequency before the median class

fm= frequency of the median class

C= size of the median class

Example 1;

The table below shows the marks obtained by forty pupils in a mathematics test

	Marks	0 – 9	10 – 19	20 – 29	30 – 39	40 – 49	50 – 59
	No. of pupils	4	5	6	12	8	5

Calculate the median of the distribution.

Solution:

	Marks	Class boundaries	(f)	Cf
	0 – 9	0 – 9.5	4	4
	10 – 19	9.5 – 19.5	5	9
	20 – 29	19.5 – 29.5	6	15
	30 – 39	29.5 – 39.5	12	27
	40 – 49	39.5 – 49.5	8	35
	50 – 59	49.5 – 59.5	5	40

Median =L1+[n2−Cfbfm]C

n2=(402)th=20th number

We find the class interval where the median lies, with the aid of the cumulative frequency 20 lies in the  after 15.  i.e class interval 30 – 39

L1=29.5, Cfb=15, fm=12, C=10 (i.e.30−20)

Median =29.5+[402−1512]10=29.5+[20−1512]10=29.5+[512]10=29.5+0.147×1029.5+4.17=33.67

Therefore, median mark =33.67

Class Activity:

1. The frequency distribution shows the marks of 100 students in a mathematics test.

	Marks	No. of students
	1 – 10	2
	11 – 20	4
	21 – 30	9
	31 – 40	13
	41 – 50	18
	51 – 60	32
	61 – 70	13
	71 – 80	5
	81 – 90	3
	91 – 100	1

Calculate the median mark.  (WAEC SSCE)

2. The table below shows the weight distribution of 40 men in a games village.

	Weight (kg)	110 – 118	119 – 127	128 – 136	137 – 145	146 – 154	155 – 163	164 – 172
	Frequency	9	3	4	5	2	5	12

Calculate the median of the distributions

The Mode of Grouped Data

Mode formula for grouped data is given as;

Mode =L1+[Δ1Δ1+Δ2]C

OR

=L1+[fxfx+fy]C

Where, L1= Lower class boundary of the modal class

Δ1 or fx= Difference between the modal frequency and the frequency of the next lower class i.e class before it

Δ2 or fy= Difference between the modal frequency and the frequency of the next highest class i.e class after it

C= Size of the modal class

Mathematics ss2 third term

Example 1:

The table below shows the weekly profit in naira from a mini-market

	Weekly profit	1 – 10	11 – 20	21 – 30	31 – 40	41 – 50	51 – 60
	Frequency	6	6	12	11	10	5

What is the modal weekly profit?

Solution:

	Weekly profit	Class boundaries	Frequency
	1 – 10	0.5 – 10.5	6
	11 – 20	10.5 – 20.5	6
	21 – 30	20.5 – 30.5	12
	31 – 40	30.5 – 40.5	11
	41 – 50	40.5 – 50.5	10
	51 – 60	50.5 – 60.5	5

The modal class is 21–30 (i.e class with the highest frequency)

Mode =L1+[Δ1Δ1+Δ2]C.

L1=20.5,Δ1=12−6=6,Δ2=12−11=1,C=10

Mode =20.5+[66+1]10=20.5+[67]1020.5+0.8571×1020.5+8.571=29.07

∴ Modal profit is ₦29.07

Example 2:

The frequency distribution of the weights of 100 participants in a women conference held in Jupiter is shown below.

	Weight (kg)	40 – 49	50 – 59	60 – 69	70 – 79	80 – 89	90 – 99	100 – 109
	No. of women	9	2	22	30	17	4	16

Calculate the modal weight of the women

Solution:

	Weight (kg)	Class boundaries	No. of women (f)
	40 – 49	39.5 – 49.5	9
	50 – 59	49.5 – 59.5	2
	60 – 69	59.5 – 69.5	22
	70 – 79	69.5 – 79.5	30
	80 – 89	79.5 – 89.5	17
	90 – 99	89.5 – 99.5	4
	100 – 109	99.5 – 109.5	16

Modal class =21−30,

L1=69.5,fx=30−22=8,fy=30−17=13,C=79.5−69.5=10

Mode =L1+[fxfx+fy]C=69.5+[88+13]×10=69.5+[821]×10=69.5+0.381×10=69.5+3.81=73.31

∴ Modal weight =73.3kg(3 s.f)

Mathematics ss2 third term

Class Activity:

The table below shows the age distributions of the members of a club.

	Age (years)	10 – 14	15 – 19	20 – 24	25 – 29	30 – 34	35 – 39
	Frequency	7	18	25	17	9	4

Calculate the modal age. (WAEC SSCE)

PRACTICE EXERCISE:

1. If 8kg of coffee costing #2000 a kg is mixed with 12kg of another kind of coffee costing #2200 a kg, what is the cost of the mixture per kg?
2. Three kinds of tea at #1,160, #1,460 and #1,540 per kg are in the ratio 2:3:5. What is the mixture worth per kg.
3. Four ingredients costing #320 per kg, #240 per kg, #160 per kg and #80 per kg are mixed so that their masses are in ratio 4:1:2:3. Calculate the average cost per kg of the mixture.
4. A trader mixes three bags of sugar costing #900/bag with seven sacks of sugar which cost #700/bag. If she sells the mixture at #950/bag, calculate her percentage profit.
5. A trader bought three kinds of nuts at #100 per kg, #84 per kg and #60 per kg respectively. He mixed them in the ratio 3:5:4 respectively and sold the mixed nuts to make a profit of 25%. At what price per kg did sell them?

ASSIGNMENT:

1. The marks scored by 30 students in a particular subject are as follows:

39732931113050335118319151413763256410761934138642269892320

(a) Prepare a frequency table, using class intervals 1 – 20, 21 – 40 e.t.c

(b) Calculate the mean mark

(c) Calculate the modal score

2. The table below shows the monthly profit in #100,000 of naira of a super market

	Monthly profit in #100,000	11 – 20	21 – 30	31 – 40	41 – 50	51 – 60	61 – 70
	Frequency	5	11	9	10	7	8

(a) What is the modal monthly?

(b) Estimate the mean and the median profit

 

Performance Metrics

Points Balance: 2

CONTENT:

(a) Definitions of: (i) Range, (ii) Variance, (iii) Standard deviation.

(b) Calculation of range, variance and standard deviation.

(c) Practical application in capital market reports; (i) Home (ii) Health studies (iii) Population studies.

 

DEFINITION AND CALCULATION OF RANGE

Measures of Dispersion

The measure of dispersion (also called measure of variation) is concerned with the degree of spread of the numerical value of a distribution.

Range:  This is the difference between the maximum and minimum values in the data.

Examples 1:

Find the range of the data 6,6,7,9,11,13,16,21 and 32

Solution: The maximum item is 32

The minimum item is 6

∴ Range =32−6=26

Example 2:

Find the range of the distributions below 65, 62, 62, 61, 61, 60, 60, 59, 58, 52

Solution:  Range =65−52=13

Deviation from the mean:

If the mean of a distribution is subtracted from any value in the distribution, the result is called the DEVIATION of the value from the mean.

Consider the table below (set of examination marks)

	65	62	62	61	61
	60	60	59	58	52

The mean =65+62+62+61+61+60+60+59+58+5210=60010=60

Deviation from the mean =x−x¯=65−60=+562−60=+2=62−60=+2=61−60=+1=61−60=+1=60−60=0 etc.

The deviations of the scores from the mean are +5, +2, +2, +1, +1, 0, 0, -1, -2, -8

The sum of these deviations =0

Class Activity:

1. Calculate the range of the following distributions

(a) 72,78,72,90,72,83,79

(b) 9,4.0,4.2,3.9,3.8,4.0

2. Calculate the mean deviation of (1a) and (1b) above

DEFINITION AND CALCULATION OF VARIANCE

The variance is the arithmetic mean of the squares of the deviation of the observations from the true mean. It is also called the mean squared deviation.

The formula for variance is

(a) ∑(x−x¯)2n, for an ordinary distribution (ungrouped)

(b) ∑f(x−x¯)2∑f, for a frequency distribution table (grouped)

Example 1:

Calculate the variance of the following distributions of the ages of 50 pupils in a secondary school

	Age (years)			10	12	13	14		15	16
	Number of pupils			18	4	6	12		6	4
	Age (x)	Freq (f)	fx					|x−x¯|			|x−x¯|2	f|x−x¯|2
	10	18	180					2.6			6.76	121.68
	12	4	48					0.6			0.36	1.44
	13	6	78					0.4			0.16	0.96
	14	12	168					1.4			1.96	23.52
	15	6	90					2.4			5.76	34.56
	16	4	64					3.4			11.56	46.24
		∑f=50	∑fx=628									∑f|x−x¯|2=228.4

Mean x¯=∑fx∑f=62850=12.56=12.6

Variance =∑f(x−x¯)2∑f=228.450=4.568≈4.6

Example 2:

Calculate the variance of the distribution below.

90, 80, 72, 68, 64, 56, 52, 48, 36, 34

Solution:

Mean x¯ or m=∑xn=60

	x	x−m=d	d2
	90	+30	900
	80	+20	400
	72	+12	144
	68	+8	64
	64	+4	16
	56	−6	16
	52	−8	64
	48	−12	144
	36	−24	576
	34	−26	676
			Total = 3000

Variance =∑d2n=300010=300

Mathematics ss2 third term

Class Activity:

Calculate the mean and variance of the ages of 12 students aged 16, 17, 18, 16.5, 17, 18, 19, 17, 17, 18, 17.5  and  16

 

Definition and Calculation of Standard Deviation

Standard deviation (S.D) is the square root of variance.

The formula for S.D are: (a) ∑(x−x¯)2n−−−−−−√ and ∑f(x−x¯)2∑f−−−−−−−√

Example 1:

Find the variance and standard deviation of the set of numbers 2,5,6,3 and 4

Solution:  Variance =∑(x−x¯)2n

But mean =4

	x	x−x¯	(x−x¯)2
	2	−2	4
	5	1	1
	6	2	4
	3	−1	1
	4	0	0
			∑(x−x¯)2=10

Variance =105=2

Standard deviation, S.D=∑(x−x¯)2n−−−−−−√=2–√=1.414

Example 2:

Calculate the standard deviation of the distribution

	Age (years)	10	12	13	14	15	16
	Frequency	18	4	6	12	6	4

Solution:

Reference to example 1 under the DEFINITION AND CALCULATION OF VARIANCE

Standard Deviation =∑f(x−x¯)2∑f−−−−−−−√=228.450−−−−√=4.568−−−−√=2.14

Class Activity:

Compute (i)  the variance  (ii)  the standard deviation of the data.

1. In a college, the number of absentees recorded over a period of 30 days was a shown in the frequency distribution table.

	Number of absentees	0 – 4	5 – 9	10 – 14	15 – 19	20 – 24
	Number of days	1	5	10	9	5

2. The table shows the distribution of ages of workers in a company

	Age (in yrs)	17 – 21	22 – 26	27 – 31	32 – 36	37 – 41	42 – 46	47 – 51	52 – 56
	Frequency	12	24	30	37	45	25	10	7

PRACTICAL APPLICATION IN CAPITAL MARKET REPORT

EXAMPLE:

1. Two groups of eight students in a class were given a test in English. Group A had the following marks; 60,70,50,48,68,72,80 and 56

Group B had the following marks: 50,90,40,58,90,82,60 and 44

(a) Calculate the mean, range, variance and standard deviation of each group.

(b) Which group had less variation in its marks?

Solution:

(a) Group A

	x	x−x¯	|x−x¯|	|x−x¯|2
	60	−3	3	9
	70	7	7	49
	50	−13	13	169
	48	−15	15	225
	68	+5	5	25
	72	+9	9	81
	80	+17	17	289
	56	−7	7	49
				∑|x -\bar{x}|^2 = 896

Mean \bar{x} = \frac{60 + 70 + 50 + 48 + 68 + 72 + 80 + 56}{8} \\ = \frac{504}{8} \\ = 63

Range = 70 -50 = 20

Variance (V) = \frac{∑(x -\bar{x})^2}{n} \\ = \frac{896}{8} \\ = 112

S.D = \sqrt{\frac{∑(x -\bar{x})^2}{n}} \\ =\sqrt{\frac{896}{8}} \\ = \sqrt{112} \\ = 10.5830 \\ = 10.58 \text{ (2 d.p)}

GROUP B:

	x	|x -\bar{x}|	|x -\bar{x}|^2
	50	14.25	203.0625
	90	25.75	663.0625
	40	24.25	588.0625
	58	6.25	39.0625
	90	25.75	663.0625
	82	17.75	315.0625
	60	4.25	18.0625
	44	20.25	410.0625
			∑|x -\bar{x}|^2 = 2899.5

Mean = 64.25

Variance = 362.43

S.D = 19.04 \text{ (2 d.p)}

(b) Group A

Class Activity:

1. The rainfall in millimetres from June to November in two towns is given below

		June	July	Aug.	Sept.	Oct.	Nov.
	Town A	1.8	2.7	1.4	2.4	2.8	1.5
	Town B	3.4	3.6	2.2	2.5	2.8	1.2

(a) Compare the means and standard deviations of rainfall in towns A and B

(b) In which town is rainfall less widely spread during the period?

2. Compute the

(i) Variance

(ii) Standard deviations

(iii) Range of the following distributions

	Score	95	85	80	75	70	65	55	40
	Frequency	1	1	1	4	1	3	1	3

 

all

CONTENT:

Histograms of grouped data (Revision): (a) Need for grouping (b) Calculation of; (i) class boundaries (ii) class interval (iii) class mark.

(b) Frequency polygon

(c) Cumulative Frequency graph: (a) Calculation of cumulative frequencies. (b) Drawing of cumulative frequency curve graph (Ogive). (c) Using graph of cumulative frequencies to estimate; (i) Median (ii) Quartiles (iii) Percentiles.(iv) Other relevant estimates. (d) Application of ogive to everyday life.

Let the record below be the mass of some people (in kg)

	66	48	71	61	39	68	33	60	52	44
	33	49	81	58	59	71	42	88	68	91
	80	66	70	26	96	63	76	46	51	61
	54	32	50	59	41	55	38	56	86	62
	50	69	23	84	77	33	71	42	69	93

Should bar chart be drawn for the different masses above, there would be too many bars, so the data may be grouped into class intervals and then a frequency distribution table prepared. Appropriate class intervals are : 21 – 30, 31 – 40, 41 – 50, …

Each data belongs to one of the class intervals. Each data is first represented by a stroke in the tally column. Every fifth stroke is used to cross the first four counted. The number of tally in each class interval gives the frequency

		Class interval		Tally		Frequency
		21 – 30	//			2
		31 – 40	////	/		6
		41 – 50	////	////		9
		51 – 60	////	////		9
		61 – 70	////	////	/	11
		71 – 80	////	/		6
		81 – 90	////			4
		91 – 100	///			3

The modal class is the one with the highest frequency.

Class Activity:

1. Prepare a frequency table, using class intervals 1 – 20, 21 – 40, … for the scores by 30 students.

	26	23	29	30	91	51
	37	64	86	9	20	19
	39	31	50	18	51	63
	33	13	31	25	41	76
	10	34	42	89	73	11

2. The marks scored by fifty students in an examination paper are given below:

	43	27	31	43	22	31	47	34	18	15
	30	45	48	55	39	25	31	12	18	21
	26	19	38	10	44	43	51	33	59	54
	41	35	37	41	46	33	51	37	48	58
	17	19	23	26	29	38	57	36	35	44

Prepare a frequency table, using class intervals 10 – 19, 20 – 29, 30 – 39, e.t.c

What is the modal class?

Mathematics ss2 third term

Calculation of (i)  class boundaries (ii) class interval (iii) class mark

Grouped data can be represented using a kind of rectangles called histogram. The width of these rectangles is determined by the class interval while the height is proportional to the frequency in that interval. To close up the gaps between the class intervals, the class interval at both ends to have a common boundary in-between two intervals. From the last frequency table above we get this table.

	Class intervals	Frequency	Class boundaries
	21 – 30	2	20.5 – 30.5
	31 – 40	6	30.5 – 40.5
	41 – 50	9	40.5 – 50.5
	51 – 60	9	50.5 – 60.5
	61 – 70	11	60.5 – 70.5
	71 – 80	6	70.5 – 80.5
	81 -90	4	80.5 – 90.5
	91 – 100	3	90.5 – 100.5

To get a common boundary between two class interval, the upper class limit of a class is added to the lower class limit of the next class and divide the sum by 2.

E.g. 20+212=412=20.530+312=612=30.5 etc.

The upper class boundary of a class is the lower class boundary of the next class. This gives a continuous horizontal axis.

Another thing to consider is the class mark or class centre. This may be used in finding the mean. For any class interval, the class center is the average of the upper and lower limits of that particular class interval.

Class center of interval 21 – 30 is 20+302=512=25.5

Class mark for class interval 31 – 40 is 31+402=712=35.5

The class mid-values (class centre) are used in plotting frequency polygon.

 

CUMULATIVE FREQUENCY GRAPH

The Cumulative frequency of a given class or group is the sum of the frequency of all the classes below and including the class itself.

Cumulative frequency curve or Ogive is a statistical graph gotten by plotting the upper class boundaries against cumulative frequencies. It is used to determine among the others: Median, Percentiles (100 divisions), Deciles (10 divisions), Quartiles (4 divisions)

Mathematics ss2 third term

The cumulative frequencies are placed along the y – axis, while the scores or class boundaries are placed along the x-axis

Calculation of cumulative frequencies and Drawing of cumulative frequency curve graph (Ogive)

Example 1;

The table below shows the frequency distributions of the lengths (in cm) of fifty planks cut by a machine in the wood-processing factory of Kara Sawmill (Nigeria)

	Class interval	21 – 30	31 – 40	41 – 50	51 – 60	61 – 70	71 – 80	81 – 90	91 – 100
	Frequency	2	6	9	9	11	6	4	3

(a) Prepare a cumulative frequency table for the distribution

(b) Draw the cumulative frequency curve (Ogive) for the distribution

Scale: 2cm to represent 10 units on the frequency axis

2cm to represent 10 units on the length axis

all

Solution:

The cumulative frequency table is given below as;

	Class interval	Class boundaries	Frequency	Cummulative frequency
	21 – 30	20.5 – 30.5	2	2
	31 – 40	30.5 – 40.5	6	6 + 2 = 8
	41 – 50	40.5 – 50.5	9	9 + 8 = 17
	51 – 60	50.5 – 60.5	9	9 + 17 = 26
	61 – 70	60.5 – 70.5	11	11 + 26 = 37
	71 – 80	70.5 – 80.5	6	6 + 37 = 43
	81 – 90	80.5 – 90.5	4	4 + 43 = 47
	91 – 100	90.5 – 100.5	3	3 + 47 = 50

To plot the graph, it is advisable to use a suitable scale. The graph should be drawn big, because the bigger the graph the more accurate the answers that would be obtained from the graph.

Cumulative frequency curve

Using graph of cumulative frequencies to estimate median, quartiles, percentiles etc

To estimate median and quartiles from the Ogive or cumulative frequency curve, we take the following steps;

STEP 1: Compute to find their position on the cumulative frequency (CF) axis using the following formulae,

(a) For lower quartile or first quartile (Q1), we use 14N

(b) For median quartile or second quartile (Q2), we use 12N

(c) For upper quartile or third quartile (Q3), we use 34N (Total frequency or last CF)

Mathematics ss2 third term

STEP 2:  Locate the point on the cumulative frequency axis and draw a horizontal line from this point to intersect the Ogive.

STEP 3: At the point it intersect the Ogive, draw a line parallel to the cumulative frequency axis to intersect the horizontal axis.

STEP 4: Read the value of the desired quartile at the point of intersection of the vertical line and the horizontal axis.

Inter-quartile range =Q3−Q1

Semi inter-quartile range =Q3−Q12

Percentile

This is the division of the cumulative frequency into 100 points. For instance;

75%=75100×N20%=20100×N

Then, we trace the required values to the graph (curve) then to the class boundaries to get the required answer.

Example 1:
The frequency distribution of the weight of 100 participants in a high jump competition is as shown below:

(a) Construct the cumulative frequency table

	Weight (kg)	20 – 29	30 – 39	40 – 49	50 – 59	60 – 69	70 –79
	No. of participants	10	18	22	25	16	9

(b) Draw the cumulative frequency curve

(c) From the curve, estimate:

(i) The median

(ii) The lower quartile

(iii) The upper quartile

(iv) The inter-quartile range

(v) The semi inter-quartile range

(vi) 65 percentile

(vii) 4^th decile

(viii) The probability that a participant chosen at random weighs at least 60kg

Solution:

	Class interval	Class boundaries	Frequency	Cummulative frequency
	20 – 29	19.5 – 29.5	10	10
	30 – 39	29.5 – 39.5	18	28
	40 – 49	39.5 – 49.5	22	50
	50 – 59	49.5 – 59.5	25	75
	60 – 69	59.5 – 69.5	16	91
	70 – 79	69.5 – 79.5	9	100

Mathematics ss2 third term

Mathematics ss2 third term

Mathematics ss2 third term

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