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6 STATISTICS Class Intervals Class Interval 1. Data that consist of the measurement of a quantity can be grouped into few classes and the range of each class in known as the class interval. Class Limits and Boundaries Lower Limit and Upper Limit 2. For class interval, for example 30 – 39, the smaller value (30) is known as the lower limit while the larger value (39) is known as theupper limit. Lower Boundary and Upper Boundary 3. The lower boundary of a class interval is the middle value between the lower limit of the class interval and the upper limit of the class before it. 4. The upper boundary of a class interval is the middle value between the upper limit of the class interval and the lower limit of the class after it. Example: 20 – 29 30 – 39 40 – 49

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Page 1: Statistics notes spm add maths

6 STATISTICSClass Intervals

Class Interval

1. Data that consist of the measurement of a quantity can be grouped into few classes and the range of each class in known as the class interval.

Class Limits and Boundaries

Lower Limit and Upper Limit2. For class interval, for example 30 – 39, the smaller value (30) is known as the lower limit while the larger value (39) is known as theupper limit.

Lower Boundary and Upper Boundary3. The lower boundary of a class interval is the middle value between the lower limit of the class interval and the upper limit of the class before it.

4. The upper boundary of a class interval is the middle value between the upper limit of the class interval and the lower limit of the class after it.

Example:

20 – 29 30 – 39 40 – 49

Class size

5. The class size is the difference between the upper boundary and lower boundary of the class.

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Example:Size of class interval 30 – 39= Upper boundary – Lower boundary= 39.5 – 29.5= 10

Mode and Mean of Grouped Data

Mode and Mean of Grouped Data

(A) Modal ClassThe modal class of grouped data is the class interval in the frequency table with the highest frequency.

(B) Class MidpointThe class midpoint is the value of data that lies at the centre of a class.

   Class midpoint=Lower limit + Upper limit2   

(C) Calculating the Mean of Grouped DataThe steps to calculate the mean of grouped data are as follows.Step 1: Calculate the midpoint value of each class.Step 2: Calculate the value of (frequency × midpoint value) of each class.Step 3: Calculate the sum of the values of (frequency × midpoint value)            of all the classes.Step 4: Calculate the sum of all the frequencies of all the classes.Step 5: Calculate the value of the mean using the formula below.

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Example:The following frequency table shows the number of magazines sold at a bookshop for 30 days in April 2013.

Number of magazines Frequency220 – 229 3230 – 239 5240 – 249 11250 – 259 6260 – 269 5

Based on the data given,(a) calculate the size of class,(b) state the modal class,(c) calculate the mean number of magazine sold per day. 

Solution:(a) Size of the class      = upper boundary – lower boundary      = 229.5 – 219.5      = 10

(b) Modal class = 240 – 249 (Highest frequency)

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(c)

Number of magazines Frequency (f) Class midpoint (x)220 – 229 3 224.5230 – 239 5 234.5240 – 249 11 244.5250 – 259 6 254.5260 – 269 5 264.5

Histograms with Class Interval of the Same SizeHistograms

Draw a histogram based on the frequency table of a grouped data

1. A histogram is a graphical representation of a frequency distribution.2. A histogram consists of vertical rectangular bars without any spacing    between them.3. Steps for drawing a histogram(a) Determine the lower boundaries and upper boundaries for each class       interval.(b) Choose suitable scales for the horizontal axis (x-axis) to represent       class interval and the vertical axis (y-axis) to represent frequency.Example:

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The following frequency table shows the radii, in cm, of different types of trees in a garden.

Radii (cm) Frequency2.0 – 2.4 72.5 – 2.9 53.0 – 3.4 103.5 – 3.9 24.0 – 4.4 64.5 – 4.9 4

Draw a histogram to represent the above information.

Solution:Radii (cm) Frequency Lower boundary Upper boundary

2.0 – 2.4 7 1.95 2.452.5 – 2.9 5 2.45 2.953.0 – 3.4 10 2.95 3.453.5 – 3.9 2 3.45 3.954.0 – 4.4 6 3.95 4.454.5 – 4.9 4 4.45 4.95

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Frequency PolygonsFrequency Polygons

1. A frequency polygon is a line graph that connects the midpoints of     each class interval at the top end of each rectangle in a histogram.

2. A frequency polygon can be drawn from a(a) Histogram,(b) Frequency table

3. Steps for drawing a frequency polygon:

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Step 1: Add a class with 0 frequency before the first class and add also a              class with 0 frequency after the last class.Step 2: Calculate the midpoints or mark the midpoints of the top sides of              the rectangular bars including the midpoints of the two              additional classes.Step 3: Joint all the midpoints with straight lines.

Example:The following frequency table shows the distance travelled by 38 teenagers by motorcycles in one afternoon.

Journey travelled (km) Frequency55 – 59 460 – 64 465 – 69 770 – 74 875 – 79 980 – 84 6

Draw a frequency polygon based on the frequency table.

Solution:

Journey travelled (km) Frequency Midpoint50 – 54 0 5255 – 59 4 5760 – 64 4 6265 – 69 7 6770 – 74 8 7275 – 79 9 7780 – 84 6 8285 – 89 0 87

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Cumulative FrequencyA) Cumulative FrequencyCumulative Frequency of a data or a class interval in a frequency table is obtained by determining the sum of its frequency with the total frequencies of all its previous data or class interval.

(B) OgiveOgive is a cumulative frequency graph which is obtained by plotting the cumulative frequency against the upper boundaries of each class.

Example:The data below shows the number of books read by a group of 60 students in a year.

Books Frequency6-10 3

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11-15 716-20 1121-25 1626-30 1131-35 836-40 4

(a) Construct a cumulative frequency table for the given data.(b) By using the scales of 2cm to 5 books on the horizontal axis and 2cm       to 10 students on the vertical axis, draw an ogive for the data. 

Solution:(a) Add a class with frequency 0 before the first class. Find the upper boundary of each class interval.

Books Frequency Cumulative Frequency

Upper Boundary

1-5 0 0 5.56-10 3 0 + 3 = 3 10.511-15 7 3 + 7 = 10 15.516-20 11 10 + 11 = 21 20.521-25 16 21 + 16 = 37 25.526-30 11 37 + 11 = 48 30.531-35 8 48 + 8 = 56 35.536-40 4 56 + 4 = 60 40.5

(b)