Download Statistics Exam 2: Probability Distributions and Confidence Intervals and more Exams Probability and Statistics in PDF only on Docsity! Page 1 of 4 Statistics 150: Introduction to Statistics Exam 2 - Chapter 5, 6, 7 October 20, 2000 Name: ____________________________________ Section 1 Directions: This section contains seven computational problems worth a total of 85 points. For each problem, you must present all your work, as well as the final answer, in order to receive full or partial credit. 1. The starting time of your favorite Sunday 2 o’clock radio show is uniformly distributed between 1:59 p.m. and 2:02 p.m. (a) Draw a diagram that represents the distribution of the starting time of this radio show. (Note: Label the axes correctly.) (7 pts.) f(x) 1/3 1:59 (0) 2:02 (3) x (b) What is the probability that the show will start before 2:00 p.m. this Sunday? (8 pts.) (1)(1/3) = 1/3 2. The net weight of “Folgers 13 oz.” is normally distributed with a mean of 468 g and a standard deviation of 3 g. You just purchased one can at a nearby grocery store. (a) What is the probability that the net weight of your can will be 465 g or greater? (8 pts.) P(X > 465) = P((X – 468)/3 > (465 – 468)/3) = P(Z > –1.00) = .3413 + .5 = .8413 (b) If your can is at the 45th percentile in the net-weight distribution, what is the actual net weight of your can? (8 pts.) P(Z < z) = .4500 ⇒ z = –0.13 –0.13 = (X – 468)/3 ⇒ x = 467.61 Page 2 of 4 3. An auto manufacturer recently discovered that 30% of the light switches installed on its particular model were defective. If the manufacturer inspects 1,000 vehicles of this model to be shipped to dealerships, what is the probability that more than 275 of them will have a defective light switch? (Note: Use the normal approximation and apply the correction for continuity.) (9 pts.) X ~ BIN(1000, .30) µ = np = 1000(.30) = 300 σ = (npq)1/2 = (1000(.30)(.70))1/2 = 14.491 Y ~ N(300, 14.1912) P(X > 275) ≈ P(Y > 275.5) = P((Y – 300)/14.491) > (275.5 – 300)/14.491) = P(Z > –1.69) = .4545 + .5 = .9545 4. Presented right is the probability distribution of the lifetime of a 1-kW mercury-vapor light bulb. The mean lifetime (µ) is 5,000 hours and the standard deviation (σ) is 5,000 hours. If one hundred 1-kW bulbs are purchased for a football stadium, what is the probability that the mean lifetime of them will exceed 6,000 hours? (9 pts.) x y 0 5000 10000 15000 20000 0. 0 0. 00 00 5 0. 00 01 0 0. 00 01 5 0. 00 02 0 The sampling distribution of X will be approximately normal (CLT). µ = 5000 σ = 5000 Xµ = 5000 Xσ = 5000/100 1/2 = 500 P( X > 6000) = P(( X – 5000)/500 > (6000 – 5000)/500) = P(Z > 2.00) = .5 – .4772 = .0228