Chapter 8 Homework ( ) -- Normal Distribution

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1 Chapter 8 Homework ( ) -- Normal Distribution Dr. McGahagan NOTE: I often abbreviate the text declaration that X is a random variable distributed normally with mean 8 and variance of 144 as " X is N (8, 12)" The N stands for "Normally distributed" and the mean and standard deviation follow in parentheses. 1. Identify clearly the area you want to find -- a sketch helps avoid confusion. Is the area specified as greater than or less than a given value, or between two values? Is the critical value above or below the mean of the distribution? If two values, are both on one side of the mean or not? 2. Normalize the critical value: subtract the distribution mean and divide by the standard deviation, so that the new critical value is standard normal -- that is, has mean zero and standard deviation one. We can call this value the z-score: Z = (critical value k - mean of distribution) / Std.dev. of distribution 3. Look up the critical value in appropriate tables such as Table A3-1 (p ) and find the corresponding area. Compare the area given by the tables to your graph (step 1 above), and shade it in on your graph. Alternative: use the (normal-cdf k) in EcLS, which will give the area under the curve up to point k, or P (Z < k). 4. If the areas shown in the graph are the same, you have the answer. If not, you must think about the relation of the two: a. The table will give you the area less than a given value; if you want to find the area greater than that value, simply subtract the area the table provides from 1.0 (the total area under the curve). P (X > k) = 1 - P (X < k) b. If you want to find the area between two points (say k1 and k2, where k2 is the larger value) look up both points in table A3 or issue the two commands (normal-cdf k2) and (normal-cdf k1), then subtract the larger area from the smaller. Note that EcLS allows you to take some shortcuts: > (p-value k) gives the probability value to the right of point k, or P (Z > k) = 1 - P(Z < k) > (normal-area k) not only gives P (Z > k), but creates a graph to show you the area visibly. > (normal-area k1 k2) gives the area between the two points k1 and k2, and creates ashaded graph. Compare the results from: (- (normal-cdf 2) (normal-cdf 1)) and (normal-area 1 2) > (normal-quant <prob>) [for example, (normal-quant.8)] gives the z-score which would result in a probability of 80 percent. See Problems 8 and 9 for examples. The Student's t distribution is very like the normal -- however, since it tries to adjust for the fact that we usually estimate the standard deviation on the basis of the data we have, it requires not only the critical value but also the degrees of freedom of the data (D.F. = number of observations minus 1). Suppose 10 observations were in our sample data; to find the probability of an observation greater than 2 standard deviations away from the mean: (- (t-cdf 2 9) 1.0) or (p-value 2 9) (t-area 2 9) will provide a graph as well as giving you the p-value; compare the results to (t-area 2 50). What happens as the degrees of freedom of the t-distribuion increase?

2 Homework -- Chapter 8. Statistics -- Dr. McGahagan Be sure to look over the conceptual questions ("Do you know the basics?" on pages 194-5) Problem 1. Annual rainfall in Johnstown is distributed normally with a mean of 52 inches and a standard deviation of 15 inches. A. What is the probability that we will get less than 45 inches of snow in a given year? i. standardize the critical value of 45 inches Z = (45-52) / 15 = - 7 / 15 = ii. sketch and shade a graph; you will be shading an area entirely to the left of the peak of the normal curve. iii. Look up P (Z < -.47) in table A3; it will be This is the desired number. or issue the command (normal-cdf ) which will give the more accurate result You should report the result as "There is about a 32 percent chance we will get less than 45 inches of snow in any given year". Note that (normal-area ) will give the mirror image of the desired area. B. What is the probability that we will get more than 40 inches of snow? i. Standardize the critical value of 40 inches. Z = (40-52) / 15 = - 12 /15 = ii. Sketch and shade a graph: this time, it will be to the right of a point below the mean, and take up more than half the area under the normal curve. iii. Look up P (Z < -.47) in table A3, it will be Shade the area: it will be the area to the left of that you shaded in step ii. iv. Subtract this area from 1.0 to get the desired area or.7881 (about 79 percent) Note that (normal-area -0.8) will get the result directly. Problem 1 in the text: Distribution of rainfall is N (40, 5) [Normal with mean 40 and SD 5] A. find P (X < 33) = P (Z < -7/5) = P(Z < -1.40) =.0808 B. find P (X > 38) = P (Z > - 2/5) = P(Z > -0.40) = 1 - P(Z < -0.40) = =.6554 Try (normal-area -0.40) to see part B graphically. Problem 2. Text problem is similar to problem 1, part B -- both call for "greater than" computations. Given normal distribution with mean 550 and SD 30 [note: we take square root of variance], A. P (X > 575) = P(Z > ( ) / 30) = P (Z > ) = 1 - P(Z <.8333) = = [using (normal-cdf.8333); table A3 forces rounding to 0.83, so would give =.2033] B. P (X > 540) = P (Z > ( ) / 30) = P (Z > ) = 1 - P(Z < ) = = [using (normal-cdf ); table A3 would give = Problem 3. Run off tackle will gain average of 2.5 yards with SD of 1 yard. Hence probability of gaining more than 4 yards is P (X > 4) = P(Z > (4-2.5) / 1) = P (Z > 1.5) = 1 - P (Z < 1.5) = 1 - P (Z < 1.5) = =.0668 (normal-area 1.5) will get the graph of this situation. Problem 4. Consider a flea-flicker play which gains an average of zero yards but has a standard deviation of 20 yards (sometimes the pass is completed, sometimes the inexperienced passer holds on until sacked). Before you choose the off-tackle play as having higher average yards, consider the odds: P (X > 4) = P(Z > 4/20) = 1 - P(Z <.2) = = The flea-flicker has a much better chance of making the required yardage than the "safe" off-tackle run. [on text problem 4: note that the end-around play is N (3, (sqrt 6)) -- you are given the variance. Hence P (X > 4) = P(Z > (4-3) / (sqrt 6) = P(Z > ) = 1 - (normal-cdf.4082) =.3416 ]

3 Problem 5. Measuring the speed of light. The first thing to do with this problem is change units from centimeters per second to tens of thousands kilometers per second, so that you can re-express the standard deviation as 5 megakilometers per second. [I know I'm creating a new metric unit here, but it makes the problem simpler]. If the expected value of your measurement is the true measurement, what is the chance that your actual measurement is within 2 megakilometers per second of the true value? P (Z < 2/5) = P(Z < -2/5) = ; subtract the two to get.3108 as your answer (using the tables) Visualization helps here: use the command (normal-area -.4.4) = and note the graph. Problem 6. Lemonade stand, with daily sales distributed N (15, 10), and hence with weekly sales distributed with a mean of 105 and a standard deviation of (sqrt 700) = The probability of selling more than 120 glasses is given by: P(X > 120) = P(Z > ( ) / (sqrt 700) = P (Z >.5669) = 1 - P(Z <.5669) = = The probability of selling more than 100 glasses is: P(X > 100) = P (Z > - 5 / (sqrt 700) = 1 - P(Z < ) = = Problem 7. Given that X is N(0,10), what is Prob (5 < X < 10)? Standardize to translate to: Pr (Z < 10/10) - Pr (Z < 5/10) = Pr (Z < 1.0) - Pr (Z < 0.5) or (normal-cdf 1.0) - (normal-cdf 0.5) = or Problem 8. If X is N(-3, 3), and A is such that its normalized score is Z = (A - (- 3)) / 3, so that (normal-cdf Z) =.6, then what is A? Using the tables, you would have to go to the body of the table rather than the margins, and find P (Z < a) =.6. The closest answers available (halfway down the first column of p.493) are and.6026, corresponding to a normalized z-score of 0.25 and [from the answer in the text, the authors chose to go with 0.25] A more precise answer is available from the normal-quant function in EcLS: (normal-quant.6) = We know that.25 = A + 3 / 3 or that.75 = A + 3 or that A = Start from this answer and work through the problem the normal way to confirm it. The Z-score of A will be ( ) / 3 =.25, and (normal-cdf.25) gives us the sixty percent probability that we were originally given. Problem 9. If X is distributed N(0, sd) -- where sd stands for the unknown standard deviation of the distribution -- and Prob (X < 5.0) =.80, what is sd? Look in the table to find 0.80 in the body of the table; at the middle of the second column on p. 493 we see that P (Z <.84) = and P (Z <.85) = Since the Z score was found by 5-0 / sd we must have 5 / sd =.84 or sd = 5 / 0.84 = The variance of the distribution must be (square ) = If you used the (normal-quant.80) command, you would have found that the most accurate Z score was.8416, so that sd = 5 /.8416 = , for a variance of Problem 10. If the mean of a distribution is 3.0, why can't the probability that ( X < 4) be 0.4? Answer: 4 is above the mean, and the standard deviation is positive, so the Z-score has to be 4-3 / sd = 1 /sd > 0. If the standard deviation is very large, the Z-score might be close to zero, but P (Z < 0) = 0.5. If the standard deviation is small (say sd = 1), P( Z < 1) = Whatever the standard deviation (try another value or three), the lowest possible value for the probability that X < 4 is 0.5

4 Problem 11. If X is N(73, 9) the probability that the absolute value of X is greater than 100 is the sum of the probabilities P (X < -100) and P (X > 100). Z-scores are ( ) / 9 = and (100-73) / 9 = 3.0, Since P (Z < ) + Pr (Z > 3.0) = (normal-cdf ) (normal-cdf 3.0) = = Problem 12. The probability that X = is zero; the probability that X is between 24.9 and 25.1 can be calculated as (normal-area ( / ( ) 10) (/ ( ) 10) ) = Problem 13. If X is N (1, 8), for what values of A is the cumulative density between 0.10 and 0.30? We can find the appropriate Z scores by looking for 0.10 and 0.30 in Table A3 (p.493) -- the Z-scores will be and To translate these back into X values, we multiply by 8 and then add on 1.0 and get and Problem 14. If the peak of the density of X is equal to 5.0 at x = 10, what are the mean and standard deviation of X? The easy part is that the mean of a normal distribution is at the peak, so mean = 10. To find the standard deviation, use the formula on p * (sqrt (2*pi)) *sd = (e 0) [since we have x - mu = 0 at x = 10, the entire exponent on e is zero] or (sqrt (2 * pi)) * sd = 1 / 5 or (2 * pi) * var = 1/25 or var = and sd = Problem 15. Given X is N(mux, sdx) and Y is N(muy,sdy) what is the probability that Y is greater than X? [See p. 182, item 5 and set c1 = 1, c2 = -1 and c3 = 0.] Construct the random variable Q = Y - X. It will have mean muq = muy - mux and (assuming X and Y are independent) variance Varx + Vary, so that sdq = (sqrt (varx + vary)) We are to find the probability that Q > 0; the normalized z-score will be Z = (0 - muq / sdq) and the answer will be 1 - (normal-cdf Z) Problem 16. The traveling salesman question implements problem 15. Let X be the random variable describing the industrial job. X is distributed N (15, 2) Let Y be the random variable describing the sales job. Y is distributed N (12, 10) Construct the variable Q representing Y - X. It will have mean = -3.0 and variance = = 104 so it will have standard deviation of To find the probability that Q > 0, construct the z-score = [0 - (- 3.0)] / =.2942 P(Q > 0) = P (Z >.2942) = 1 - P (Z <.2942) = 1 - (normal-cdf ) =.3843 [Using table A3, you would get 1 - P (Z <.29) = =.3859, which the text answer rounds to.39]

5 Problem 17. A bank knows that its daily deposits are normally distributed with mean 12 (thousand dollars) and standard deviation 4 (thousand dollars). Its daily withdrawals are distributed N (10, 5). We will want to create random variable Q = Withdrawals minus Deposits, which will have mean = -2 and variance = = 41, so that the standard deviation of Q is (sqrt 41) = To find the probability that: a. Deposits for a given day are more than 13 (thousand dollars). P (D > 13) = P (Z > (13-12) / 4) = P (Z > 0.25) = 1 - P (Z < 0.25) = =.4013 b. Withdrawals for a given day are more than 13 (thousand dollars). P (W > 13) = P (Z > (13-10) / 5) = P (Z > 0.6) = 1 - P (Z < 0.6) = = c. Withdrawals exceed deposits: P (W - D > 0) = P (Q > 0) = P (Z > (0 - (- 2)) / (sqrt 41))) = P (Z >.3123) = 1 - P (Z <.3123) = [Text differs slightly because it uses 1 - P (Z <.31) = =.3783] d. Withdrawals are 5 (thousand dollars) greater than deposits: P ( W - D > 5) = P (Q > 5) = P (Z > (5 - (- 2)) / (sqrt 41)) = P (Z > ) = 1 - P (Z < ) = 1 - (normal-cdf ) = = Problem 18. Adult heights are distributed N (68 inches, 5 inches) in a town with 10,000 adults. What is the expected number of people who are: a. Taller than 6 foot 6 = 78 inches. P (X > 78) = P (Z > 10 / 5) = 1 - P (Z < 2) = so * 10,000 = 227 and one half people. Since for half a person to be 6 foot six, the whole person would have had to be 13 feet tall, the text did wisely to report 227 people. b. Between 6 feet and 6 foot 6 (72 to 78 inches). Calculate the Z-scores: z1 = / 5 = 4/5 =.80 z2 = / 5 = 2.0 P (Z < 2.0) =.9772 P (Z < 0.8) =.7881 The difference is.1891, and the expected number of adults in this range is.1891 * 10,000 = 1,891 c. Between 5 feet 8 and 6 feet tall? That is, from 68 inches to 72 inches? Z1 = / 5 = 0 P (Z1 < 0) = 0.5 Z2 = / 5 = 0.8 P (Z2 <.8) = for a difference of which should translate into 2,881 people in that range. d. Between 5 feet 4 inches = 64 inches and 5 feet 8 inches = 68 inches. Z2 = / 5 = 0 P (Z < 0) = 0.5 Z1 = / 5 = P (Z < - 0.8) = so =.2881 of the population are in this range; that is 2,881 people. Since we have been looking at people in a range of 4 inches above and below the mean, it should not be surprising that this answer is the same as that in part c. e. Shorter than 5 feet 4 inches = 64 inches? Z = / 5 = - 0.8; P(Z < - 0.8) =.2119, so 2,119 people should be shorter than this. Problem 19. Circuit lifetime is N (50, 8) where 50 = 50,000 hours mean time before failure. a. The probability that any one of these circuits will last less than 30 (thousand hours) is: P (X < 30) = P (Z < / 8) = P (Z < - 20/8) = P (Z < -2.5) = b. The probability that it will last more than 55 (thousand hours) is: P (X > 55) = P (Z > / 8) = 1 - P (Z < 5/8) = =.2660 [The text used 0.62 for 5/8, and hence found P (Z < 0.62) =.7327]

6 Problem 20. Banquet arrivals expected to be 65, plus or minus 4, which translates into banquet arrivals being distributed N (65, 4). What is the number of places to set if you want the probability of an overflow to be under 3 percent? P (X > N) <.03, where X is the number of arrivals, and N (to be chosen) the number of place settings. Translate into the standard normal distribution: P (Z > (N - 65) / 4) <.03 or 1 - P (Z < (N - 65) / 4) <.03 or 0.97 < P (Z < (N - 65) / 4) We can first find the term N - 65 / 4 by looking for = 0.97 in the body of the text table A3 The appropriate value is 1.86, so N - 65 / 4 = 1.88 or N - 65 = 7.52 or N = Round up to 73 place settings if you do not want to break the fine china plates in half. Problem 21. Log normal income distribution. In this case the text uses log to the base 10. The average income value cited is , and if 10 is raised to the power, the result will be an income of $ 20,000. The following table will help with the problem: X (income) Log X Z-score ( X ) / 0.3 $ 10, $ 15, $ 20, $ 25, $ 30, Enabling you to calculate (for example), the probability of an income above $ 25,000 as 1 - P (Z < ( ) / 0.3)) = 1 - P (Z < ) = = Problem 23. Using other distributions is similar to using the normal distribution. To find the probability that a variable distributed chi-square with 3 d.f. is greater than 4, calculate 1 - (chisq-cdf 4 3) = = Problem 24. The variable of interest, X, is distributed χ 2 (5 df). The probability that its value is less than 2 is given by (chisq-cdf 2 5) = Note that the text answers to 24 and 25 are only approximations (and not very good ones) on the basis of table A3-3. Use the computer, not tables, to answer these questions. Problem 25. If X is distributed Student's t with 5 df, the probability that it is between 0 and 1 is: (t-cdf 1 5) - (t-cdf 0 5) = = Problem 26. (t-cdf -2 10) - (t-cdf -1 10) = = [Again, the text answers based on table A3-4 are not very good approximations] Problem 27. Given a variable distributed χ 2 (12 df), how likely is an observation greater than or equal to 16? Use (chisq-area 16 12) to get the appropriate graph and answer ( or 19 percent). This is no more surprising than snow in winter in Johnstown. Problem 28. Given a variable distributed χ 2 (16 df), how likely is an observation less than or equal to 5? Use (chisq-area 5 16) and find the area of the unshaded portion; or use (chisq-cdf 5 16) to get Not very likely (like snow in Los Angeles) but it happens sometimes, so the text answer "don't believe" is not necessarily to be given in all circumstances. Problem 29. Given a variable distributed Student's t with 3 df, how likely is an observation of or less?use (t-area ) and compute the unshaded area or use (t-cdf ) = Problem 30. How likely is an observation of 0.7 or more given a t-distribution with 12 degrees of freedom? (t-area ) gives the answer of or about 25 percent.