My Math Forum Which of the two sets of data intervals has the greater standard deviation?

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 July 24th, 2017, 09:56 PM #1 Member   Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 Which of the two sets of data intervals has the greater standard deviation? A company is evaluating two employees' times to complete the same repetitive task. Employee A's time and frequency Under two minutes: 2 2 to 3.99 minutes: 6 4 to 5.99 minutes: 9 6 to 7.99 minutes: 2 8 to 9.99 minutes: 1 Employee B's time and frequency Under two minutes: 3 2 to 3.99 minutes: 5 4 to 5.99 minutes: 5 6 to 7.99 minutes: 4 8 to 9.99 minutes: 3 Which employee has the greater standard deviation and why?
 July 25th, 2017, 06:00 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Okay, do you know what "standard deviation" means? For the first set of data the mean is (2(1)+ 6(3)+ 9(5)+ 2(7)+ 9(5)+ 2(6)+ 9)/(2+ 6+ 9+ 2+ 1)= 100/20= 5. So the variance is (2(1- 5)^2+ 6(3- 5)^2+ 2(7- 5)^2+ (9- 5)^2)/20= (2(16)+ 6(4)+ 2(4)+ 16)/20= (32+ 24+ 8+ 16)/20 and the standard deviation is the square root of that.
July 25th, 2017, 06:48 AM   #3
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 Originally Posted by Country Boy Okay, do you know what "standard deviation" means? For the first set of data the mean is (2(1)+ 6(3)+ 9(5)+ 2(7)+ 9(5)+ 2(6)+ 9)/(2+ 6+ 9+ 2+ 1)= 100/20= 5. So the variance is (2(1- 5)^2+ 6(3- 5)^2+ 2(7- 5)^2+ (9- 5)^2)/20= (2(16)+ 6(4)+ 2(4)+ 16)/20= (32+ 24+ 8+ 16)/20 and the standard deviation is the square root of that.
In this problem the actual means cannot be determined, since the data for each employee is given in intervals (2 to 3.99 minutes, etc.) rather than exact times. So I think this problem relies on just looking at data in the chart for each employee to determine which set has the greater standard deviation.

Assuming the middle interval is "4 to 5.99 minutes," then Employee A has a frequency of 9 (11 other data points which deviate). Employee B has a frequency of 5 in the same interval (15 other data points that deviate). So my thought is that Employee B's set of data has the greater standard deviation but I'm not sure.

 July 25th, 2017, 02:33 PM #4 Member   Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 I assigned actual values to the data at the lower and upper values in each interval Employee A's time and frequency Under two minutes: 2 2 to 3.99 minutes: 6 4 to 5.99 minutes: 9 6 to 7.99 minutes: 2 8 to 9.99 minutes: 1 low end of each interval 0, 0, 2 , 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6, 6, 8, The mean would be 68/20 = 3.4 (lowest possible mean) high end of each interval 1.99, 1.99, 3.99, 3.99, 3.99, 3.99, 3.99, 3.99, 5.99, 5.99, 5.99, 5.99, 5.99, 5.99, 5.99, 5.99, 5.99, 7.99, 7.99, 9.99, The mean would be 107.8/20 = 5.39 (approx. highest possible mean) Employee A data's standard deviation for lowest possible mean of 3.4: (3.4-0)^2+(3.4-0)^2+(3.4-2)^2+(3.4-2)^2+(3.4-2)^2+(3.4-2)^2+(3.4-2)^2+(3.4-2)^2+(3.4-4)^2+(3.4-4)^2+(3.4-4)^2+(3.4-4)^2+(3.4-4)^2+(3.4-4)^2+(3.4-4)^2+(3.4-4)^2+(3.4-4)^2+(3.4-4)^2+(3.4-6)^2+(3.4-6)^2+(3.4- 8 )^2 = 72.8 SQUARE ROOT (72.8/20) = 1.908 Standard deviation of Employee A's data If using the upper end of the data (mean 5.39), then the standard deviation doesn't change. (5.39-1.99)^2+(5.39-1.99)^2+(5.39-3.99)^2+(5.39-3.99)^2+(5.39-3.99)^2+(5.39-3.99)^2+(5.39-3.99)^2+(5.39-3.99)^2+(5.39-5.99)^2+(5.39-5.99)^2+(5.39-5.99)^2+(5.39-5.99)^2+(5.39-5.99)^2+(5.39-5.99)^2+(5.39-5.99)^2+(5.39-5.99)^2+(5.39-5.99)^2+(5.39-7.99)^2+(5.39-7.99)^2+(5.39-9.99)^2 = 72.8 still a standard deviation of 1.908 --------------------------------------------------------------------------------------- Employee B's time and frequency Under two minutes: 3 2 to 3.99 minutes: 5 4 to 5.99 minutes: 5 6 to 7.99 minutes: 4 8 to 9.99 minutes: 3 The low end of each interval: 0, 0, 0, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8 The mean would be 78/20 = 3.9 Employee B data's standard deviation for lowest possible mean of 3.9: (3.9-0)^2+(3.9-0)^2+(3.9-0)^2+(3.9-2)^5+(3.9-2)^5+(3.9-2)^5+(3.9-2)^5+(3.9-2)^5+(3.9-4)^2+(3.9-4)^2+(3.9-4)^2+(3.9-4)^2+(3.9-4)^2+(3.9-6)^2+(3.9-6)^2+(3.9-6)^2+(3.9-6)^2+(3.9- 8 )^2+(3.9- 8 )^2+(3.9- 8 )^2 = 131.8 SQUARE ROOT (131.8/20) = 2.567 Standard deviation of Employee B's data Employee A's data has a standard deviation of 1.908 Therefore, Employee B's data has the greater standard deviation. Last edited by Seventy7; July 25th, 2017 at 02:43 PM.
 July 25th, 2017, 03:22 PM #5 Member   Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 Another way without too many calculations: Employee A's time and frequency Under two minutes: 2 2 to 3.99 minutes: 6 4 to 5.99 minutes: 9 6 to 7.99 minutes: 2 8 to 9.99 minutes: 1 as 1.99, 1.99, 3.99, 3.99, 3.99, 3.99, 3.99, 3.99, 4, 4, 4, 4, 4, 4 ,4 ,4 ,4 ,6, 6, 8 The overall mean would be 4.196. 15 data points are relatively close to that mean, with only 5 deviating a greater amount. Employee B's time and frequency Under two minutes: 3 2 to 3.99 minutes: 5 4 to 5.99 minutes: 5 6 to 7.99 minutes: 4 8 to 9.99 minutes: 3 Can try to minimize the standard deviation by adjusting the data in the intervals: 1.99, 1.99, 1.99, 3.99, 3.99, 3.99, 3.99, 3.99, 4, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8 But 10 of the data (bolded) are still relatively further away from the mean than Employee A's data, especially the two extra 8's. Therefore, Employee B's data has the greater standard deviation.

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