1. Advanced Math Topics 6.4 Computational Formula for Variation and Standard Deviation

2. Variance: σ 2 = Σ(x – μ) 2 p(x) Do you remember the variance formula?

3. A bowling ball manufacturer makes bowling balls in 2 pound intervals from 8 to 18 pounds. The probability that a customer will buy a particular weighted ball is shown. Find the mean, variance, and standard deviation. x (lbs.)p(x) x p(x)x - μ(x – μ) 2 (x – μ) 2 p(x) 80.11 0.888 – 12.6 = -4.621.16(21.16)(0.11) = 2.3276 100.21 2.1010 – 12.6 = -2.66.76(6.76)(0.21) = 1.42196 120.28 3.3612 – 12.6 = -0.60.36(0.36)(0.28) = 0.1008 140.17 2.3814 – 12.6 = 1.41.96(1.96)(0.17) = 0.3332 160.13 2.0816 – 12.6 = 3.411.56(11.56)(0.13) = 1.5028 180.10 1.8018 – 12.6 = 5.429.16(29.16)(0.10) = 2.916 σ 2 = Σ(x – μ) 2 p(x) μ = 12.6 σ 2 = 8.60236 σ = √8.60236 ≈ 2.9330 Do you remember this example?

4. A bowling ball manufacturer makes bowling balls in 2 pound intervals from 8 to 18 pounds. The probability that a customer will buy a particular weighted ball is shown. Find the mean, variance, and standard deviation. x (lbs.)p(x) x p(x)x 2 x 2 p(x) 80.11 0.8864(64)(0.11) = 7.04 100.21 2.10100(100)(0.21) = 21 120.28 3.36144(144)(0.28) = 40.32 140.17 2.38196(196)(0.17) = 33.32 160.13 2.08256(256)(0.13) = 33.28 180.10 1.80324(324)(0.10) = 32.4 σ 2 = Σx 2 p(x) – μ 2 μ = 12.6 167.36 σ = √8.6 ≈ 2.9326 There is a computational formula that will give you the same answer, despite a possible rounding difference. σ 2 = 167.36 – (12.6) 2 = 8.6

5. 3) Janet is a medical lab technician. The number of EEG’s that she takes daily and the associated probabilities are shown. Find the mean, variance, and standard deviation for the distribution. From the HW P. 306 x (EEG’s)p(x) 00.08 10.09 20.13 30.07 40.14 50.23 60.19 70.07 μ = 3.9 σ 2 = 19.25 – (3.9) 2 = 4.29 σ = √4.29 ≈ 2.07 sum = 19.5

6. P. 306 #3, 12, and 13; for #12 and 13, compute each using both formulas; Quiz tomorrow From the HW P. 306