Statistics — Practice

Question 1

Find the range of the data: 12, 45, 8, 33, 27, 51, 19.

Accepted Answer

Maximum = 51, minimum = 8. Answer: Range = 51 - 8 = 43.

Question 2

Find the mean deviation about the mean for: 2, 4, 5, 7, 8.

Accepted Answer

Mean = (2+4+5+7+8)/5 = 26/5 = 5.2. Absolute deviations: 3.2, 1.2, 0.2, 1.8, 2.8, summing to 9.2. Answer: Mean deviation = 9.2/5 = 1.84.

Question 3

Find the mean deviation about the median for: 10, 12, 8, 15, 9, 14, 11.

Accepted Answer

Arranged: 8,9,10,11,12,14,15 (7 values, odd), median is the 4th value = 11. Absolute deviations from 11: 3,2,1,0,1,3,4, summing to 14. Answer: Mean deviation = 14/7 = 2.

Question 4

Find the variance and standard deviation of: 6, 8, 10, 12, 14.

Accepted Answer

Mean = 50/5 = 10. Deviations: -4,-2,0,2,4; squared: 16,4,0,4,16, summing to 40. Variance = 40/5 = 8. Answer: Standard deviation = sqrt(8) = 2.sqrt(2), approximately 2.83.

Question 5

Find the variance of the first 6 natural numbers.

Accepted Answer

Data: 1,2,3,4,5,6. Mean = 21/6 = 3.5. Deviations squared: 6.25,2.25,0.25,0.25,2.25,6.25, summing to 17.5. Answer: Variance = 17.5/6, approximately 2.92.

Question 6

Using the step-deviation method with assumed mean A=25 and h=10, find the mean and variance for: xi = 5,15,25,35,45 with fi = 3,7,10,6,4.

Accepted Answer

yi = (xi-25)/10 = -2,-1,0,1,2. fi.yi = -6,-7,0,6,8, summing to 1. fi.yi^2 = 12,7,0,6,16, summing to 41. N=30. Mean = 25 + 10(1/30) = 25.33. Variance = (100/900)[30(41)-1^2] = (100/900)(1229) = 136.56. Answer: Mean is approximately 25.33 and variance is approximately 136.56.

Question 7

The mean and standard deviation of 10 observations are 12 and 3 respectively. If every observation is multiplied by 4, find the new mean and new standard deviation.

Accepted Answer

Using the scaling property: new mean = 4(12) = 48. New variance = 4^2(3^2) = 16(9) = 144, so new SD = sqrt(144) = 12. Answer: New mean = 48 and new standard deviation = 12.

Question 8

The mean and standard deviation of 15 observations were found to be 12 and 3. Later, an observation recorded as 10 was found to actually be 16. Find the corrected mean and standard deviation.

Accepted Answer

Original sum = 15(12)=180. Corrected sum = 180-10+16=186. Corrected mean = 186/15 = 12.4. Original sum of squares = 15(3^2+12^2) = 15(153) = 2295. Corrected sum of squares = 2295-100+256=2451. Corrected variance = 2451/15 - 12.4^2 = 163.4-153.76=9.64. Answer: Corrected mean = 12.4 and corrected standard deviation = sqrt(9.64), approximately 3.10.

Practice — Statistics

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