, 0 x < 2. a. Find the probability that the text is checked out for more than half an hour but less than an hour. = (1/2)2

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1 Math 205 Spring 206 Dr. Lily Yen Midterm 2 Show all your work Name: 8 Problem : The library at Capilano University has a copy of Math 205 text on two-hour reserve. Let X denote the amount of time the text is actually checked out, and suppose the cdf is 0, x < 0 x F (x) = 2 4, 0 x < 2, 2 x Use the cdf to answer the following. a. Find the probability that the text is checked out for more than half an hour but less than an hour. P ( 2 < X < ) = F () F ( 2 ) = 2 4 (/2)2 4 = 4 6 = 6. b. Find the median check out duration. Score: / If F (x) = x2, then = (and 0 x < 2), so x2 = 2, so x = ± 2, but 0 x < 2, so x = 2. Score: / c. Find the probability density function of X, that is, f(x). 0, x < 0 f(x) = F x (x) = 2, 0 x < 2 0, x > 2 (The value of f(2) cannot be determined from the given information since F (2) is not defined. However, the value of a probability density function at a single point is immaterial.) d. Find the expected value and standard deviation of X. E(X) = xf(x) dx = 2 x x dx = x 2 = 8 0 = Therefore V (X) = (x 4 )2 f(x) dx = (x 4 0 )2 x dx = (x2 8x + 6) x dx = x 4 x2 + 8x dx = 9 8 x4 4 9 x + 4x2 2 = = 2, so σ = V (X) = 2/9 = 2.

2 Problem 2: Identify each missing value in the following normal curves. Provide 4-decimal place accuracy unless the answer is expressed by a fraction exactly. µ = 5, σ = /2 Standard normal z = µ = 0, σ = µ = 0, σ = x = z.8 0 x x Problem : Suppose 90% of all B.C. drivers regularly wear a seat belt while operating a vehicle. A random sample of 500 drivers is selected. Find the following probabilities. a. Between 00 and 50 (inclusive) of the drivers in the sample regularly wear a seat belt. This is a binomial experiment with n = 500 and p = Since np = and n( p) = 50 0, you may use the normal approximation to the binomial distribution with µ = np = 450 and σ = np( p) = 45, so P (00 X 50) = B(50; 500, 0.9) B(299; 500, 0.9) normalcdf(299.5, 50.5, 450, 45) = Using the binomial distribution yields b. Fewer than 450 of those in the sample regularly wear a seat belt. P (X < 450) = B(449; 500, 0.9) normalcdf(, 449.5, 450, 45) = Using the binomial distribution yields Score: / Page 2 Math 205

3 Problem 4: The time taken by a randomly selected applicant for filling out a form for Capilano University student loan has a normal distribution with mean value 0 minutes and standard deviation 2 minutes. If five students fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most minutes? Here µ = 0 and σ = 2. Let X, X 2, X, X 4, and X 5 be random variables measuring the time taken by each student on the first day, and let X be their average. Similarly, let Ȳ be the average on the second day. Assuming that all the students are independent, P ( X and Ȳ ) = P ( X ) P (Ȳ ) = 2 normalcdf(,, 0, 5 2 ) normalcdf(,, 0, 6 ) = = Problem 5: Given a random sample X, X 2,..., X n from a symmetric distribution with mean µ and standard deviation σ, circle all the unbiased estimators of µ. trimmed mean x tr(0) ; midrange; sample mean x; mode, and median x. Of the unbiased estimators you circled, which is the best estimator? Why? The mean, x, is the minimum variance unbiased estimator, thus the best. Page Math 205

4 Problem 6: The joint and marginal pmf s of X and Y are partly given in the following table. X \ Y 2 Total Total 6 a. Complete the table with exact values. 2 b. Find E(Y ). E(Y ) = y y P Y (y) = = = = 4 6 = 7. c. Find Cov(X, Y ). Since E(X) = = 2 and E(XY ) = x,y xyp(x, y) = = 59, it follows that Cov(X, Y ) = E((X µ X )(Y µ Y )) = E(XY ) µ X µ Y = =. 2 4 d. Are X and Y independent? Since the covariance is non-zero, X and Y are not independent. Some of you used p(x, y) = p X (x) p Y (y) for all x, y in the domain from the definition of independence. Then you checked, say, p(, ) = /6, but p X () p Y () = / /6, thus not equal, and x, y, not independent. Page 4 Math 205

5 Problem 7: A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (seconds) to complete the escape: Assume the random variable X for escape time measured in seconds is normally distributed. Answer the following. a. Compute sample mean and sample standard deviation. Using STATS CALC :-Var Stats yields that x = 70.7 s and s = 24.6 s. Here you have a sample, and must take the sample standard deviation. b. Calculate an upper confidence bound for population mean escape time using a confidence level of 95%. Use a t-distribution with µ, σ unknown, n = 26 < 40. The upper confidence bound is s x + t 0.05,25 26 = 70.7 s s 26 = 78.9 s You need to remember that when you compute a one-sided confidence bound, α is not halved. Score: / Problem 8: The following is a summary of data analysis on the girth, in centimetres, of grizzly bears for a test of H 0 : µ = 00 cm versus H a : µ 00 cm: One-Sample Z: Girth Test of µ = 00 vs µ 00 N Variable Mean Standard Deviation Z P 6 Girth a. What is the conclusion if you test with α = 0.02? Since 0.08 < 0.02, there is sufficient evidence to reject the null to conclude that the mean girth of grizzly bears is not 00 cm. If you consider only the left rejection region, then you need to halve your P-value and say 0.09 < 0.0. Many used the test statistic z = 2.7 < z 0.0 = to reject the null. Score: / b. Identify the type of mistake you could have made in part a). Rejecting the null hypotesis when it is true is a Type I error. If your conclusion in the first part is not rejecting the null, then you needscore: to be / consistent, and answer a Type II error instead. c. Before data was actually collected, what was the probability of making the mistake in part a)? α = 0.02 Score: / d. Suppose the rejection region is (, 92) (08, ) when in fact the true mean girth is µ = 90 cm. Find the probability of a Type II error. A Type II error is not rejecting the null hypothesis when it is actually false, thus, the test statistic is not in rejection region. X N(90, (2.79/ 6) 2 ), so β(90) = P (92 x 08) when µ = 90. So, β(90) = normalcdf(92, 08, 90, 2.79/ 6) = If you used the formula for two-tailed Type II error, then you must convert the region to a relevant α value first: α = , thus α/2 = β(90) = Φ(z / ) Φ( z / ) = Remember that Φ is our normalcdf. Page 5 Math 205

6 Problem 9: A sample of 50 lenses used in eyeglasses yields a sample mean thickness of.05 mm and a sample standard deviation of 0.4 mm. The desired true average thickness of such lenses is.20 mm. Does the data strongly suggest that the true average thickness of such lenses is something other than what is desired? Test using α = a. State null hypothesis. H 0 : µ =.20 mm. Please remember to include units in your statements and conclusions. b. State alternative hypothesis. H a : µ.20 mm. Score: / c. State test statistic and compute. z = x µ 0 s/.05 mm.20 mm = n 0.4 mm/ = Since α = 0.05, z α/2 =.960. Thus z =.20 <.960, so reject the null hypothesis. Score: / d. State conclusion. There is sufficient evidence to reject the claim that the true average thickness of such lenses is.20 mm. Score: / Page 6 Math 205