Part4 Review solutions

# Part4 Review solutions - Review of Part IV 279 Review of...

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Review of Part IV 279 Review of Part IV 1. Quality Control. Construct a Venn diagram of the disjoint outcomes. a) P (defect) = P (cosm.) + P (func.) – P (cosm. and func.) = 0.29 + 0.07 – 0.02 = 0.34 Or, from the Venn: 0.27 + 0.02 + 0.05 = 0.34 b) P (cosm. and no func.) = P (cosm.) – P (cosm. and func.) = 0.29 – 0.02 = 0.27 Or, from the Venn: 0.27 (the region inside Cosmetic circle, yet outside Functional circle) c) P P P (func. cosm.) func. cosm.) (cosm.) = =≈ (. . . 002 029 0 069 From the Venn, consider only the region inside the Cosmetic circle. The probability that the car has a functional defect is 0.02 out of a total of 0.29 (the entire Cosmetic circle). d) The two kinds of defects are not disjoint events, since 2% of cars have both kinds. e) Approximately 6.9% of cars with cosmetic defects also have functional defects. Overall, the probability that a car has a cosmetic defect is 7%. The probabilities are estimates, so these are probably close enough to say that they two types of defects are independent. 2. Workers. Organize the counts in a two-way table. a) i) P female) == 90 150 06 ii) PP P P (( ( ( . female production) female) production) - female and production) ∪= + =+−= 90 150 117 150 72 150 09 iii) Consider only the production row of the table. There are 72 women out of 117 production workers. 72/117 ≈ 0.615. Or, use the formula: P P P (female production) (female production) (production) = 72 150 117 150 0 615 . Male Female Total Management 7 6 13 Supervision 8 12 20 Job Type Production 45 72 117 Total 60 90 150

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280 Part IV Randomness and Probability iv) Consider only the female column. There are 72 production workers out of a total of 90 women. 72/90 = 0.8. Or, use the formula: P P P (production female) (production female) (female) = == 72 150 90 150 08 . b) These data suggest that holding a production position may be associated with gender. 60% of the plant employees are women, but 61.5% of the production workers are women. However, this is a small difference, and may be due to sampling error. 3. Airfares. a) Let C = the price of a ticket to China Let F = the price of a ticket to France. Total price of airfare = 3 C + 5 F b) µ =+ = + = + = EC F EC EF () ( ) ( ) ( ) ( ) \$ 3 5 3 5 3 1000 5 500 5500 σ = + = + SD C F Var C Var F ( ) (( ) ) ) ) ( ) ( ) \$. 3 5 3 5 3 150 5 100 672 68 2 2 2222 c) =− = = −= ( ) ( ) \$ 1000 500 500 = + SD C F Var C Var F ( ) ( ) 150 100 180 28 22 d) No assumptions are necessary when calculating means. When calculating standard deviations, we must assume that ticket prices are independent of each other for different countries but all tickets to the same country are at the same price. 4. Bipolar. Let X = the number of people with bipolar disorder in a city of n = 10,000 residents. These may be considered Bernoulli trials. There are only two possible outcomes, having bipolar disorder or not having bipolar disorder. Psychiatrists estimate that the probability that a person has bipolar is about 1 in 100, so p = 0.01. We will assume that the cases of bipolar disorder are distributed randomly throughout the populations. The trials are not independent, since the population is finite, but 10,000 people represent fewer than 10% of all people. Therefore, the number of people with bipolar disorder in a city of 10,000 may be modeled by Binom (10000, 0.01).
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Part4 Review solutions - Review of Part IV 279 Review of...

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