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S1 Measures of Dispersion The mean, variance and standard deviation.

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Presentation on theme: "S1 Measures of Dispersion The mean, variance and standard deviation."— Presentation transcript:

1 S1 Measures of Dispersion The mean, variance and standard deviation

2 S1 Measures of Dispersion Objectives: To be able to find the variance and standard deviation for discrete data To be able to find the variance and standard deviation for continuous data

3 A factory worker counted the numbers of nuts in a packet. The results are shown in the table Number of nuts Freq CF Calculate P 80 and P 40 P 80 = 99/100x80 = 79.2 P 80 = 80th term P 80 = 10 nuts P 40 = 99/100x40 = 39.6 P 40 = 40th term P 40 = 8 nuts Calculate the 40 th to 80 th interpercentile range 40 th to 80 th interpercentile range = = 2 nuts

4 The lengths of a batch of 2000 rods were measured to the nearest cm. The measurements are summarised below. Length (nearest cm) Number of rods Cumulative frequency Q1= x Q1=77.06 Q2= x Q2=81.57 Q3= x Q3=85.88 By altering the formula slightly can you work out how to find the 3 rd decile (D3) and the 67 th percentile (P67)?

5 Answers D3= x D3=78.09 P67= x P67=84.26

6 Variance of discrete data The deviation (difference) of the data (x) from the mean (x) is one way of measuring the dispersion (spread) of a set of data. _ Variance = Σ(x – x)² n _ Variance = Σx² – Σx ² n n _

7 As variance is measured in units ² you usually take the square root. This gives you the standard deviation. Standard deviation. Standard deviation = variance Standard deviation symbol is σ

8 The marks scored in a test by seven students are 3,4,6,2,8,8,5. Calculate the variance and standard deviation. x x² Σx = 36Σx² = 218 Variance = Σx² – Σx ² n n _ σ ² = 218 – 36 ² = σ = 4.69 = 2.17

9 The variance and standard deviation from a frequency table fx Number of rods (x) Freq of rods (f) fx² Σf=109 Σfx=4096Σfx²= Variance = Σfx² – Σfx ² Σf Σf Variance = – 4096 ² σ ²= σ = 1.109

10 Variance and standard deviation from a grouped frequency distribution Height of plant (cm) Freq (f) 0 < h 54 5 < h < h < h < h < h 301 Total σ ² = – 285 ² = σ = σ = 11.35


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