Bowen EHS Inc. CSP Question 04/12/2004 - Statistics PDF

1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] CSP Question 04/12/2004 - Statistics An industrial hygienist is performing an exposure assessment for toluene in a shoe factory. The IH picks a random sample of 5 employees. Their sampling times and the...

1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] CSP Question 04/12/2004 - Statistics An industrial hygienist is performing an exposure assessment for toluene in a shoe factory. The IH picks a random sample of 5 employees. Their sampling times and the average concentrations over the sampling times are given below. The TLV for toluene is 50 ppm. What are the 95% confidence limits for the average employee exposure? Employee Time(hours) Concentration (ppm) 1 6 05 2 7 30 3 8 45 4 7 10 5 4 25 A) (3.5, 31.7) B) (5.8, 29.4) C) (8.9, 37.1) D) (11.2, 34.8) (Solution on next page) Copyright © 2004 Bowen EHS, Inc.  1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] Solution: The average concentration is given by: n ∑x (5 + 30 + 45 + 10 + 25) = 23. 1 1 X = i = n i =1 5 The standard deviation is given by: n ∑ (x i =1 i − X )2 (5 − 23) 2 + (30 − 23) 2 + (45 − 23) 2 + (10 − 23) 2 + (25 − 23) 2 s= = = 16.05 n −1 5 −1 The 95% confidence limits of the average are given by: s 16.05 X ± 1.96 = 23 ± 1.96( ) n 5 16.05 16.05 23 − 1.96( ) = 8.9 23 + 1.96( ) = 37.1 5 5 The correct answer is C. Copyright © 2004 Bowen EHS, Inc.  1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] CSP Question 01/11/2007 – Statistics The average test score for students taking a certain safety exam is 163. The standard deviation of the test scores is 24. Assume the test scores are normally distributed, and estimate the test score of the 5th percentile. A. 9 B. 116 C. 124 D. 210 (Solution on next page) Copyright © 2007 Bowen EHS, Inc.  1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] Solution: Use the z-table to determine the number of standard deviations below the mean that provides the 5th percentile. z = 50% - 5% = 45% The z-value with 45% of the area under the normal distribution is 1.64. The 5th percentile is given by the mean subtracting 1.64 standard deviations. 5th percentile = 163 – 1.64(24) = 124. The correct solution is C. Copyright © 2007 Bowen EHS, Inc.  1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] CSP Question 02/10/2010 – Statistics Use the table below to determine the Spearman Rank Correlation of safety audit scores and injury recordable rate. A high audit score indicates that the location has a good safety program. A high injury recordable rate indicates the location has a lot of occupational injuries. LOCATION AUDIT SCORE RECORDABLE RATE Chicago 84 0.55 St. Louis 93 0.68 New Orleans 63 0.48 New York 88 0.52 Seattle 77 0.49 A. -1.00 B. -0.90 C. +0.90 D. +1.00 (Solution on next page) Copyright © 2010 Bowen EHS, Inc.  1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] Solution: Step 1: Set up a table with rank of audit score and recordable rate. Remember a high recordable rate means a high injury rate. Add an additional column to note the difference (D) in rank between the 2 scores. Location Audit Rank Recordable Rank Difference (D) Chicago 3 4 1 St. Louis 1 5 4 New Orleans 5 1 4 New York 2 3 1 Seattle 4 2 2 Step 2: Calculate the Spearman Rank Correlation. rs = 1- 6∑ D2 = 1− ( 6 12 + 42 + 42 + 12 + 22 )= 1− 228 = 1 − 1.9 =−0.9 ( N N2 -1 ) ( ) 5 52 -1 120 The correct solution is B. An audio visual presentation of this solution with is available for purchase at our bookstore in the audio visual category. www.bookstore.bowenehs.com Copyright © 2010 Bowen EHS, Inc.  B ow e n E H S, I nc. 1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] CSP Question 03/03/2011 – Statistics Calculate the Spearman Rank Correlation of Audit Rank and Performance Rank for the four sites. Site Audit Rank Performance Rank AA 1 4 BB 2 3 CC 3 2 DC 4 1 A. -1.00 B. -0.50 C. 0.00 D. 1.00 (Solution on next page) Copyright © 2011 Bowen EHS, Inc.  B ow e n E H S, I nc. 1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] Solution: Site Audit Rank Performance D Rank (Audit – Performance) AA 1 4 -3 BB 2 3 -1 CC 3 2 1 DC 4 1 3 rs = 1 − ( ) 6∑ D2 N (N − 1) 2 rs = 1− ( 6 ( −3)2 + ( −1)2 + 12 + 32 )= 1− 6 ( 20 ) 1− = 120 −1 = ( 4 42 − 1 ) 4 (15 ) 60 The correct solution is A. Copyright © 2011 Bowen EHS, Inc.  1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] CSP Question 04/05/2012 – Statistics Use the Poisson distribution to estimate the probability of a school having four children diagnosed with a CNS cancer in one year. The school has a population of 500, and the incidence of CNS cancer in children is approximately 21 per million children. A. 4.6 x 10^-10 B. 2.1 x 10^-5 C. 0.0105 D. 0.0402 (Solution on next page) Copyright © 2012 Bowen EHS, Inc.  1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] Solution: λ x e-λ P(x) = x! The expected frequency λ = population x frequency = 500 x (21/1,000,000) = 0.0105 0.0105 4 e-0.0105 1.09x10-8 P(4) = = = 4.6 x 10-10 4! 24 The correct solution is A. Copyright © 2012 Bowen EHS, Inc.  1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] CSP Question 11/15/2012 – Statistics Examination of a large corporation’s incident records spanning the period from January 1, 2009, to December 31, 2011, shows that contact with sharp objects was responsible for an average of 612 injuries per calendar year. During calendar year 2011, the corporation employed 29,473 full time equivalent employees who worked a combined total of 56,971,309 hours. Each employee worked an average of 1,933 hours over the year. Based on this information and assuming no major changes in employment, scheduling, equipment, methods, or other factors in the coming calendar year, what is the probability within any one particular hour of operation, that an employee will sustain an injury due to contact with a sharp object? A. 2.15 B. 3.16 x 10-1 C. 2.08 x 10-2 D. 1.07 x 10-5 (Solution on next page) Copyright © 2012 Bowen EHS, Inc.  1289 Fordham Blvd, Suite 411 Chapel Hill, NC 27514 866-264-5852 [email protected] Solution: 612 𝑠ℎ𝑎𝑟𝑝 𝑜𝑏𝑗𝑒𝑐𝑡 𝑖𝑛𝑗𝑢𝑟𝑖𝑒𝑠 𝑃(𝑖𝑛𝑗𝑢𝑟𝑦 𝑑𝑢𝑒 𝑡𝑜 𝑠ℎ𝑎𝑟𝑝 𝑜𝑏𝑗𝑒𝑐𝑡) = = 1.07 × 10−5 56,971,309 ℎ𝑜𝑢𝑟𝑠 𝑤𝑜𝑟𝑘𝑒𝑑 Likelihood is a synonym for probability. Note that we are determining a probability for a given unit of time and not determining an OSHA incident rate, which is a value that is normalized for comparison between firms. The approach to assigning probabilities based on historical data is referred to as the “relative frequency method”. To determine the relative frequency probability of an event, we divide the numbers of events that have occurred by the total number of opportunities for the event to occur; since our goal is to determine the hourly probability for the corporation as a whole, we divide by the number of hours worked by the employees of the corporation. The correct answer is D. Copyright © 2012 Bowen EHS, Inc.