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2.1 Frequency distribution Ogive. Example: page 44, problem 25 6065686366 6769636562 6473615067 7162586569 67726163 maxmin.

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Presentation on theme: "2.1 Frequency distribution Ogive. Example: page 44, problem 25 6065686366 6769636562 6473615067 7162586569 67726163 maxmin."— Presentation transcript:

1 2.1 Frequency distribution Ogive

2 Example: page 44, problem 25 6065686366 6769636562 6473615067 7162586569 67726163 maxmin

3 -range = 73 – 50 = 23 -c.w. = 23 ÷ 6 = 3.8 → 4 -we will start the setup of the class limits with the minimum data value (50), to which we will add the class width (c.w.) -these are the lower limits (where each class begins)

4 Class limits TallyfClass boundary Cumu- lative freq. 50 – 54 – 58 – 62 – 66 – 70 –

5 -Next we determine the first class’ upper limit (where it ends) by subtracting 1 from the second class’ lower limit -54 – 1= 53 -then we add again the class width (4) to the upper class’ limits -tally

6 Class limits TallyfClass boundary Cumula- tive freq. 50 – 53/1 54 – 570 58 – 61////4 62 – 65//// 9 66 – 69//// //7 70 – 73///3

7 -To find the class boundary we will subtract a half a unit (usually.5) from the lower class limits and add a half a unit (usually.5) to the upper class limits

8 Class limits TallyfClass boundary Cumula- tive freq. 50 – 53/149.5 – 53.5 54 – 57053.5 – 57.5 58 – 61////457.5 – 61.5 62 – 65//// 961.5 – 65.5 66 – 69//// //765.5 – 69.5 70 – 73///369.5 – 73.5

9 -We obtain the cumulative frequency by adding to the class’ frequency the frequencies of the classes above

10 Class limits TallyfClass boundary Cumula- tive freq. 50 – 53/149.5 – 53.51 54 – 57053.5 – 57.50 + 1 = 1 58 – 61////457.5 – 61.54 + 1 = 5 62 – 65//// 961.5 – 65.59 + 5 = 14 66 – 69//// //765.5 – 69.57 + 14 = 21 70 – 73///369.5 – 73.53 + 21 = 24

11 OGIVE -Is an open line -Always rises (at least it is horizontal) -On the x-axis: class boundary -On the y-axis: cumulative frequency -Important: label the axis, name your graph

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