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Measures of Dispersion…

Measures of Dispersion…. Prepared by: Bhakti Joshi Date: December 05, 2011. Problem 1.

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Measures of Dispersion…

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  1. Measures of Dispersion… Prepared by: Bhakti Joshi Date: December 05, 2011

  2. Problem 1 In a final exam, a student “A” obtained 70 marks in Business Statistics and 60 marks in foundation course which were compared with the entire class. The average marks and standard deviation of the entire class was 75 marks and 5 marks respectively in Business Statistics and average marks and standard deviation in foundation course was 56 marks and 2 marks respectively. Find by how much does the student “A” deviates from the class. Did student “A” perform above or below average in both the subjects?

  3. x  x _ Solution: Calculate the Standard Score Standard Score = S • Standard Score in Business Statistics = (70-75)/5 = -1 • Standard Score in Foundation Course = (60-56)/2 = +2 • Student “A” performed below average in Business Statistics (70<75) and above average in Foundation course (60>56)

  4. Solution: In terms of Standard Scores • Negative standard score means below average • Zero standard score means at the average • Positive standard score means above average Interpretation • Suppose, for an observation that shows wage of Rs 17000, the standard score calculated was - 0.6, with average wages equal to Rs 20,000 and the standard deviation was 5000. To determine by how many rupees the observation Rs 17,000 deviates from the average wage, you multiply standard score 0.6 by standard deviation 5000 and obtain minus 3000. This implies for an observation with wages Rs 17000, wages deviate less by Rs 3000.

  5. Chebyshev’s Theorem • Analyzing locations of observations around the mean in accordance to standard deviations • 68% of values in a normal distribution will fall between +1 and -1 standard deviations from the mean • About 95% of the values will fall between +2 and -2 standard deviations from the mean • About 99% of the values will fall between +3 and -3 standard deviations from the mean • No matter what the shape of the distribution, at least 75% of the values will fall between +2 and -2 standard deviations from the mean and at least 89% of the values will lie within +3 and -3 standard deviations from the mean

  6. About Normal Distribution • A normal distribution of a set of data which follows a bell shaped curve • Normally distributed curve means that the data has less tendency of producing unusually extreme values and the curve is symmetric • Mean is at the centre and the data falls evenly on either side of the mean • The spread of the normal distribution is controlled by the standard deviation. The smaller the standard deviation, the more concentrated the data • Mean and median are the same for normal distribution

  7. Normal Distribution Curve

  8. Problem 3 A set of 60 observations has a mean of 66.8, a variance of 12.60 and an unknown distribution shape. Between what values should at least 75% of the observations fall according to Chebyshev’s Theorem?

  9. Solution According to Chebyshev’s Theorem, to find the values, calculate the interval (Mean minus Standard Deviation multiplied by 2, Mean plus Standard Deviation multiplied by 2) • (66.8 – 2*3.55, 66.8 + 2*3.55) • (59.7, 73.9) : 75% of the values lie between 59.7 and 73.9

  10. Problem 4 Continuing problem 3, if the distribution is symmetrical and bell-shaped, approximately how many observations should be found in the interval 59.7 and 73.9? • The interval 59.7 and 73.9 is nothing but 66.8 – 2*3.55, 66.8 + 2*3.55, with standard deviation as 2 • According to Chebyshev’sTheorem, at least 1 – (1/k)^2 of the measurements will fall within (Mean minus Standard Deviation multiplied by 2, Mean plus Standard Deviation multiplied by 2) • i.e. 1 – (1/2)^2 = 0.75 • Thus 75% of the values will fall between 59.7 and 73.9 with mean of 66.8, a variance of 12.60

  11. Problem 5 Each day in a laboratory, technician A completes an average of 40 analyses with a standard deviation of 5, whereas technician B completes an average of 160 analyses with a standard deviation of 15. Which technician shows less variability?

  12. _ x Solution S * (100) • Coefficient of variation = • Coefficient of variation for Technician A = (5/40 )*100 = 12.5% • Coefficient of variation for Technician B = (15/160) *100 = 9.4% What does this mean?

  13. Problem In a sample of 30 carpet looms, the cumulative frequency distribution of production levels (in yards) is given as follows: How many looms made less than 16 yards of carpets?

  14. Problem Following is the frequency and cumulative distribution of fish caught by Mr. X during 20 fishing trips made through his boats. If Mr. X believes that break-even catch of his boats is 5000 pounds per trip, roughly, by what proportion of the trips breaks even for Mr. X

  15. Email: bhaktij@gmail.comWebsite: www.headscratchingnotes.net

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