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Histograms

-The AREA of each bar represents the frequency

-Sometimes a key like the red one in the histogram below is used

We will now construct a histogram to represent the following grouped continuous data. REMEMBER, discrete data is data that can only take particular values (usually integers) like "the number of boys in a class". Continuous data is data that can take any value at all like height.

HEIGHT OF PLANT FREQUENCY FREQUENCY DENSITY
0≤x<2 4 2
2≤x<4 7 3.5
4≤x<8 5 1.25

We need to work out the "frequency density" in order to draw the bars correctly on the diagram, otherwise the bars would be disproportionate when working with different widths. (Getting one person between 1&2 meters tall is not the same as getting one person between 1&1.2 meters tall. The later 'deserves' a higher bar because the interval was smaller.

Frequency density = Frequency / Class width                                   Example: 7/(4-2) = 3.5 and 5/(8-4) = 1.25

This formula comes from the fact that the area of each bar represents the frequency so height of bar ("frequency density") times the width of the bar = the frequency.

*Just a note... When dealing with ages remember that they always round down. When someone is 13 they are not 14 but 13!*

 

Averages

Mean

-Take the mid-point of each interval and add them together

-Divide by the sum of the frequencies

Example...

Value

Frequency

0≤x<6 7
6≤x<12 2
12≤x<18 6

 

Mode

-The modal interval/class is the one with the highest frequency density (the tallest bar)

 

Median

-The middle value

-It cuts histograms in half so that there is an equal area on either side

The total number of plants sampled is 16 so the median height will be the 8th one along.

How tall is the 8th one along?...

There are 4 plants in the first bar so this 8th one is NOT in the first bar.

The next bar contains the next 7 plants. Therefore the 8th plant that we are looking for is the 4th plant out of these seven. It therefore lies 4 sevenths of the way across this bar. This bar starts at a plant height of 2 cm and goes up to 4cm. Therefore the plant we are interested in lies 4 sevenths of the way between 2 and 4. This value is 2+(4/7 x 2) = 3 1/7cm

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