Chapter 3. Probability

Chapter 3: Probability 879. Acceptance Sampling 2500 aspirin tablets, 2% are defectiveP(at least one defect)= 1 – P(none of the four samples are defective)=1 – [P(first is not defective) ∗ P(second is not defective) ∗…∗ P(fourth is not defective)]=1 – 0.98 4 = 1 – 0.9224= 0.077610. Testing a Claim 12 couples have babies, P(all are girls)121= 4096P(all 12 girls)= = 0.5 12 = 0.00024412Yes, the probability of this outcome happening by chance is only about 2 times out of 10,000 and so this is notlikely a chance or random outcome.11. Selecting Members 10 members of boarda. P(three wealthiest members selected)First find number of possible selections in the form of combinations of 3 things selected from 10 things,order is not important so we find 10 C 310! 10 ∗ 9 ∗8∗7! 10 ∗ 9 ∗8720= == =(10 − 3)!3! 7!3! 3∗2 ∗1610C3=120There is only one way out of 120 samples that the three wealthiest can be selected1 =120P(three wealthiest out of 10 are selected)= 0. 00833b. Different slates of three officers to be elected from 10 membersorder is important since the officer positions are different, so we find 10 P 310! 10 ∗9∗8∗ 7!10P3= == 10 ∗9∗8= 720(10 − 3)! 7!12. Life Insurance One year policies to 12 men, 99.82% (p= 0.9982) chance of living through yearP(all survive the year)= P(first will survive) ∗ P(second will survive) ∗…∗ P(twelfth will survive) = 0.9982 12 =0.97913. Chlamydia Rate Rate was 278.32 per 100,000 populationnumber out of 100,000100,000278.32100,000a. P(selected person has Chlamydia)= = = 0. 0027832b. P(two selected have Chlamydia)= P(first has Chlamydia) ∗ P(second has Chlamydia)=0.002783 ∗ 0.002783= 0.00000775c. P(two selected don’t have Chlamydia)= P(first doesn’t) ∗ P(second doesn’t)=0.997217 ∗ 0.997217= 0.9944413Cumulative Review Exercises1. Treating Chronic Fatigue Syndrome=∑ Xxn84 =21a. Mean= = 4. 00b. Median= middle score since odd number of score= 11 th score out of 21= 4.00c. Standard deviation (to find standard deviation and variance, lets first find the sum of squares SS)22n() − ( )Sum of Squares =∑x∑xn21(430) − (84)=2129030 − 7056 1974== = 94.0021 21