# Statistics II - Mathematics Form 4 Notes

## Introduction

• These statistical measures are called measures of central tendency and they are mean, mode and median.

## Mean using Working (Assumed) Mean

• Assumed mean is a method of calculating the arithmetic men and standard deviation of a data set.
• It simplifies calculation.

Example
The masses to the nearest kilogram of 40 students in the form 3 class were measured and recorded in the table below. Calculate the mean mass
 Mass kg 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Number ofemployees 2 0 1 2 3 2 5 6 7 5 3 2 1 1
Solution
We are using assumed mean of 53
 Mass x kg t= x - 53 f ft 474849505152535455 56 57 58 5960 -6-5-4-3-2-10 12 3 4 5 67 2 0 1 2 3 2 5 67 5 3 2 11 -120 -4 -6 -6 -2 0 61415121067 Σf = 40 Σft = 40

Mean of t = Σft= 40   = 1
Σf    40
Mean of x = 53 + mean of t
= 53 + 1
= 54

## Mean of Grouped Data

• The masses to the nearest gram of 100 eggs were as follows
 Mass 100- 103 104- 107 108- 111 112-115 116-119 120-123 Frequency 1 15 42 31 8 3
Find the mean mass

Solution
Let use a working mean of 109.5.
 class Mid-point x t= x - 109.5 f ft 100-103 104-107 108-111 112-115 116-119120-123 101 .5 105.5 109.5 113.5 117.5121 .5 -8 -4 0 4 8 12 1 15 42 31 8 3 –8–6001246436 Σf=100 Σft=156
Mean of t = 156 = 1.56
100
Therefore,mean of x = 1 09.5 + mean of t
= 109.5 + 1.56
= 111.06 g
• To get the mean of a grouped data easily, we divide each figure by the class width after substracting the assumed mean.
• Inorder to obtain the mean of the original data from the mean of the new set of data, we will have to reverse the steps in the following order;
• Multiply the mean by the class width and then add the working mean.

Example
The example above to be used to demonstrate the steps
 class Mid-point x t= x-109.5        4 f ft 100-103 104-107 108-111 112-115 116- 119120-123 101.5 105.5 109.5 113.5 117.5121.5 -2 -1 0 1 2 3 1 15 42 31 8 3 –2–15031169 Σf=100 Σft=39
t=  ∑ft  = 39
∑f     100
= 0.39
Therefore ̅x = 0.39 x 4 + 109.5
= 1.56 + 109.5
= 111.06 g

## Quartiles, Deciles and Percentiles

• A median divides a set of data into two equal part with equal number of items.
• Quartiles divides a set of data into four equal parts.The lower quartile is the median of the bottom half.The upper quartile is the median of the top half and the middle coincides with the median of the whole set od data
• Deciles divides a set of data into ten equal parts. Percentiles divides a set of data into hundred equal parts.
Note;
• For percentiles, deciles and quartiles the data is arranged in order of size.

Example 1

 Height in cm 145-149 150-154 155- 159 160- 164 165-169 170-174 175-179 Frequency 2 5 16 9 5 2 1

Calculate the;

1. Median height
1. Lower quartile
2. Upper quartile
2. 80th percentile

Solution

1. There are 40 students. Therefore, the median height is the average of the heights of the 20th and 21st students.
 class frequency Cumulative frequency 145 - 149 150 - 154 155 - 159 160 - 164165 - 169170 - 174175 - 179 2 5 16 952 1 272332373940
Both the 20th and 21 st students falls in the 155 -159 class. This class is called the median class.
Using the formula m = L +(n/2 - C)i
f
Where:
L is the lower class limit of the median class
N is the total frequency
C is the cumulative frequency above the median class
I is the class interval
F is the frequency of the median class

Therefore;

Height of the 20th student = 154.5 + 13 x 5
16
= 154.5 + 4.0625
=158.5625

Height of the 21st = 154.5 + 14 x 5
16
=154.5 + 4.375
=158.875
Therefore median height = 158.5625 + 158.875
2
= 158.7 cm
1. lower quartile Q1= L +(n/4-C)i
f
The 10th student falls in the in 155 – 159 class

Q1= 154.5 +(40/4 -7)5
16
= 154.5 + 0.9375
= 155.4375
2. Upper quartile Q3= L +n -C)i
f
The 10th student fall in the in 155 – 159 class
Q3= 159.5 +  x 40 -23)5
9
= 159.5 + 3.888
= 163.3889

Note;
The median corresponds to the middle quartile Q2 or the 50th percentile
2. 80 x 40 = 32 the 32nd student falls in the 160 - 164 class
100
The 80th percentile= L + ((80/100)n − C)i
f
= 159.5 + (32 − 23)5
9
= 159.5 + 5
= 164.5

Example 2

Determine the upper quartile and the lower quartile for the following set of numbers
5, 1 0 ,6 ,5 ,8, 7 ,3 ,2 ,7 , 8 ,9

Solution

Arranging in ascending order
2, 3, 5,5,6, 7,7,8,8,9,1 0
The median is 7
The lower quartile is the median of the first half, which is 5.
The upper quartile is the median of the second half, which is 8.

### Median from Cumulative Frequency Curve

• Graph for cumulative frequency is called an ogive.
• We plot a graph of cumulative frequency against the upper class limit.

Example
Given the class interval of the measurement and the frequency, we first find the cumulative frequency as shown below.
Then draw the graph of cumulative frequency against upper class limit
 Arm Span (cm) Frequency (f) CumulativeFrequency 140 ≤ x < 145 3 3 145 ≤ x < 150 1 4 150 ≤ x < 155 4 8 155 ≤ x < 160 8 16 160 ≤ x < 165 7 23 165 ≤ x < 170 5 28 170 ≤ x < 175 2 30 Total 40
Solution

Example
The table below shows marks of 1 00 candidates in an examination
 Marks 1 -10 11 -20 21 -30 31 -40 41 -50 51 -60 61 -70 71 -80 81 -90 91 -100 Frequency 4 9 1 6 24 1 8 1 2 8 5 3 1
1. Determine the median and the quartiles
2.  If 55 marks is the pass mark, estimate how many students passed
3. Find the pass mark if 70% of the students are to pass
4. Determine the range of marks obtained by
1. The middle 50 % of the students
2. The middle 80% of the students

Solution
 Marks 1 -10 11 -20 21 -30 31 -40 41 -50 51 -60 61 -70 71 -80 81 -90 91 -100 Frequency 4 9 16 24 18 12 8 5 3 1 cumulative frequency 4 13 29 53 71 83 91 96 99 100

Solution

1. Reading from the graph
The median = 39.5
The Lower quartile Q1 = 28.5
The upper quartile Q2 = 53.5
2. 23 candidates scored 55 and over
3. Pass mark is 31 if 70% of pupils are to pass
4.
1. The middle 50% include the marks between the lower and the upper quartiles i.e. between 28.5 and 53.5 marks.
2.  The middle 80% include the marks between the first decile and the 9th decile i.e between 18 and 69 marks

## Measure of Dispersion

### Range

• The difference between the highest value and the lowest value

• It depends only on the two extreme values

#### Interquartile range

• The difference between the lower and upper quartiles. It includes the middle 50% of the values

#### Semi quartile range

• The difference between the lower quartile and upper quartile divided by 2.
• It is also called the quartile deviation.

### Mean Absolute Deviation

• If we find the difference of each number from the mean and find their mean , we get the mean Absolute deviation

### Variance

• The mean of the square of the square of the deviations from the mean is called is called variance or mean deviation.

Example

 Deviation from mean(d) +1 -1 +6 -4 -2 -11 +1 10 fi 1 1 36 1 6 4 121 1 100

Sum d2 = 1 + 1 + 36 + 16 + 4 + 121 + 1 + 100 = 280
Variance =∑d2280= 35
N         8

The square root of the variance is called the standard deviation. It is also called root mean square deviation.
For the above example its standard deviation = √35 = 5.9

Example

The following table shows the number of children per family in a housing estate

 Number of children 0 1 2 3 4 5 6 Number of families 1 5 11 27 10 4 2

Calculate

1. The mean number of children per family
2. The standard deviation

Solution

 Number of children (x) Number of Families (f) fx Deviations d= x - m d2 fd2 0 1 2 3 4 5 6 1 5 11 27 104 2 0 5 22 81 402012 – 3 – 2 -1 0 1 2 3 9 4 1 0 1 4 9 920110101618 Σf=60 ∑fx=180 Σfd2=–40
1. Mean =∑fx180 = 3 childen
Σf       60
2. Variance = ∑fd2
∑f
=84
60
= 1.4
standard deviation = √1.4 = 1.183

Example

The table below shows the distribution of marks of 40 candidates in a test

 Marks 1 -10 11- 20 21 -30 31 -40 41 -50 51 -60 61 -70 71 -80 81 -90 91 -100 frequency 2 2 3 9 1 2 5 2 3 1 1

Calculate the mean and standard deviation.

 Marks Midpoint( x) Frequency(f) fx d= x - m d2 fd2 1 - 10 11 - 20 21 - 30 31 - 40 41 - 50 51 - 60 61 - 7071 - 8081 - 9091 - 1 00 5.5 15.5 25.5 35.5 45.5 55.5 65.575.585.595.5 2 2 3 9 1  5 2 3 1 1 11 .0 31 .0 76.5 319.5 546.0 277.5 131 .0226.585.595.5 - 39.5 -29.5 -19.5 -9.5 0.5 10.5 20.530.540.550.5 1560.25 870.25 380.25 90.25 0.25 110.25 420.25930.251640.252550.25 3120.51740.51140.75812.253.00551.25840.52790.751640.252550.25 Σf=40 Σfx=1800 Σfd2= 15190

Mean ̅x = ∑fx = 1800= 45 marks
∑f       40
Variance = ∑fd2= 15190= 379.75
∑f         40
= 379.8
Standard deviation = √379.8 = 19.49
Note;
- Adding or subtracting a constant to or from each number in a set of data does not alter the value of the variance or standard deviation.

## More Formulas

• The formula for getting the variance s2 of a variance x is
d2 = ∑(x-̅x)2f
∑f
= ∑fx(∑fx/∑f)2
∑f

Example

The table below shows the length in centimeter of 80 plants of a particular species of tomato

 length 152-156 157-161 162-166 167-171 172- 176 177-181 frequency 12 14 24 15 8 7

Calculate the mean and the standard deviation

Solution

Let A = 169

 Length Mid-point x x-169 t= x-169        5 f ft ft2 152 - 156 154 -15 -3 12 -36 108 157 - 161 159 -10 -2 14 -28 56 162 - 166 164 -5 -1 24 -24 24 167 - 171 169 0 0 15 0 0 172 - 176 174 5 1 8 8 8 177 - 181 179 1 0 2 7 14 28 ∑f=80 ∑ft=-66 ∑ft2=224

̅t = -66 = - 0.825
80
Therefore ̅x = -0.825 x 5 + 169
= -4.125 + 169
= 164.875 ( to 4 s.f)
Variance of t =
∑ft2 t2
∑f

= 224 - (-0.825)2
80

= 2.8 – 0.6806
= 2.119
Therefore, variance of x = 2.119 x
52
= 52.975
= 52.98 ( 4 s.f)
Standard deviation of x = √
52.98
= 7.279= 7.28 (to 2 d.p)

## Past KCSE Questions on the Topic

1. Every week the number of absentees in a school was recorded. This was done for 39 weeks these
observations were tabulated as shown below
 Number of absentees 0.3 4 -7 8 -1 1 1 2 - 1 5 1 6 - 1 9 20 - 23 (Number of weeks) 6 9 8 1 1 3 2
Estimate the median absentee rate per week in the schoo
2. The table below shows high altitude wind speeds recorded at a weather station in a period of 100 days.
 Wind speed (knots) 0 - 19 20 - 39 40 - 59 60-79 80- 99 100-119 120-139 140-159 160-179 Frequency (days) 9 1 9 22 18 13 1 1 5 2 1
1. On the grid provided draw a cumulative frequency graph for the data
2. Use the graph to estimate
1. The interquartile range
2. The number of days when the wind speed exceeded 1 25 knots
3. Five pupils A, B, C, D and E obtained the marks 53, 41, 60, 80 and 56 respectively. The table below shows part of the work to find the standard deviation.
 Pupil Mark x x - a ( x-a)2 A B C D E 5341608056 -5-17222-2
1. Complete the table
2. Find the standard deviation
4.  In an agricultural research centre, the length of a sample of 50 maize cobs were measured and recorded as shown in the frequency distribution table below.
 Length in cm Number of cobs 8 – 1011 – 1314 – 1617 – 1920 – 2223 - 25 4 7 111585
Calculate
1. The mean
2.
1. The variance
2. The standard deviation
5. The table below shows the frequency distribution of masses of 50 new- born calves in a ranch
Mass (kg)           Frequency
15 – 18                2
19 – 22                3
23 – 26               10
27 – 30               14
31 – 34               13
35 – 38               6
39 – 42               2
1. On the grid provided draw a cumulative frequency graph for the data
2. Use the graph to estimate
1. The median mass
2. The probability that a calf picked at random has a mass lying between 25 kg and 28 kg.
6. The table below shows the weight and price of three commodities in a given period
Commodity         Weight       Price Relatives
X                         31               25
Y                         41               64
Z                         21               40
Calculate the retail index for the group of commodities.
7. The number of people who attended an agricultural show in one day was 51 0 men, 1 080 women and some children. When the information was represented on a pie chart, the combined angle for the men and women was 21 60. Find the angle representing the children.
8. The mass of 40 babies in a certain clinic were recorded as follows:
Mass in Kg           No. of babies.
1.0 – 1.9              6
2.0 – 2.9              14
3.0 – 3.9              10
4.0 – 4.9              7
5.0 – 5.9              2
6.0 – 6.9              1
Calculate
The inter – quartile range of the data.
The standard deviation of the data using 3.45 as the assumed mean.
9. The data below shows the masses in grams of 50 potatoes
 Mass (g) 25-34 35-44 45-54 55-64 65-74 75-84 85-94 No of potatoes 3 6 16 12 8 4 1
1. On the grid provide, draw a cumulative frequency curve for the data
2. Use the graph in (a) above to determine
1. The 60th percentile mass
2. The percentage of potatoes whose masses lie in the range 53g to 68g
10. The histogram below represents the distribution of marks obtained in a test.

The bar marked A has a height of 3.2 units and a width of 5 units. The bar marked B has a height of 1.2 units and a width of 10 units
If the frequency of the class represented by bar B is 6, determine the frequency of the class represented by bar A.
11.  A frequency distribution of marks obtained by 1 20 candidates is to be represented in a histogram.
The table below shows the grouped marks. Frequencies for all the groups and also the area and height of the rectangle for the group 30 – 60 marks.
 Marks 0-10 10-30 30-60 60-70 70-100 Frequency 12 40 36 8 24 Area of rectangle 180 Height of rectangle 6
1.
1. Complete the table
2. On the grid provided below, draw the histogram
2.
1. State the group in which the median mark lies
2. A vertical line drawn through the median mark divides the total area of the histogram into two equal parts. Using this information or otherwise, estimate the median mark
12. The table shows the number of bags of sugar per week and their moving averages
 Number of bags per week 340 330 x 343 350 345 Moving averages 331 332 y 346
1. Find the value of X and Y axis
2. Find the order of the moving average

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Read 15594 times Last modified on Monday, 17 January 2022 12:27

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