2. Closing prices for two stocks were recorded on ten successive Fridays. Mean = 61.5 Median = 62 Mode = 67 Mean = 61.5 Median = 62 Mode = 67 56 57 58 61 63 67 Stock AStock B 33 42 48 52 57 67 77 82 90

3. Range for A = 67 – 56 = $11 Range = Maximum value – Minimum value Range for B = 90 – 33 = $57 The range is easy to compute but only uses two numbers from a data set.

4. The deviation for each value x is the difference between the value of x and the mean of the data set. In a population, the deviation for each value x is: In a sample, the deviation for each value x is: Deviations:

5. – 5.5 – 4.5 – 3.5 – 0.5 1.5 5.5 56 57 58 61 63 67 56 – 61.5 57 – 61.5 58 – 61.5 Stock A Deviations:

6. – 5.5 – 4.5 – 3.5 – 0.5 1.5 5.5 56 57 58 61 63 67 56 – 61.5 57 – 61.5 58 – 61.5 Stock A The sum of the deviations is always zero. Note: Deviations:

7. Sum of squares – 5.5 – 4.5 – 3.5 – 0.5 1.5 5.5 x 56 57 58 61 63 67 30.25 20.25 12.25 0.25 2.25 30.25 188.50 Population Variance: The sum of the squares of the deviations, divided by N. x –  (x –  

8. Population Standard Deviation: The square root of the population variance. The population standard deviation is $4.34. (Makes the units make sense)

9. Sample Variance and Standard Deviation To calculate a sample variance, s 2,divide the sum of squares by n – 1. The sample standard deviation, s, is found by taking the square root of the sample variance.

10. Summary Range = Maximum value – Minimum value PopulationSample Variance Standard Deviation

11. What is standard deviation?   

12. Computational Shortcut Sample S.D.:Population S.D.:

13. DefinitionComputational Variance Standard Deviation Use both the defining formula and the computational shortcut to calculate the variance and standard deviation.

14. Coefficient of Variation: Sample Coefficient of Variation: Population Coefficient of Variation:

15. For k = 3, at least 1 – 1/9 = 8/9 = 88.9% of the data lie within 3 standard deviations of the mean. For a distribution of any shape, the fraction of data lying within k standard deviations of the mean is at least 1 – 1/k 2 For k = 2, at least 1 – 1/4 = 3/4 = 75% of the data lie within 2 standard deviations of the mean. Chebyshev’s Theorem:

16. A survey of mountain lions finds the mean life span to be 27 years with a standard deviation of 4 years. Of a population of 1500 mountain lions, what is the minimum number you can expect to live between 19 and 35 years?

17. Chebyshev’s Theorem: A survey of mountain lions finds the mean life span to be 27 years with a standard deviation of 4 years. Of a population of 1500 mountain lions, what is the minimum number you can expect to live between 19 and 35 years?  2 standard deviations  above or below the mean

18. Chebyshev’s Theorem: A survey of mountain lions finds the mean life span to be 27 years with a standard deviation of 4 years. Of a population of 1500 mountain lions, what is the minimum number you can expect to live between 19 and 35 years?  2 standard deviations  above or below the mean Chebyshev: At least must lie within k standard deviations of the mean

19. Chebyshev’s Theorem: A survey of mountain lions finds the mean life span to be 27 years with a standard deviation of 4 years. Of a population of 1500 mountain lions, what is the minimum number you can expect to live between 19 and 35 years?  2 standard deviations  above or below the mean Chebyshev: At least must lie within k standard deviations of the mean