Chapter 6_Tutorial_W24_solution

Department of Mathematics and Statistics STAT2910-01: Statistics for Sciences Faculty of Science University of Windsor Tutorial for Chapter 6 with solution Exercise 1. Identify the choice that best completes the statement or answers the question. 1. If Z is a standard normal random variable, the area between z = 0 . 0 and z = 1 . 20 is 0 . 3849, while the area between z = 0 . 0 and z = 1 . 40 is 0 . 4192. What is the area between z = –1 . 20 and z = 1 . 40? a. 0.0343 b. 0.0808 c. 0.1151 d. 0.8041 2. If X is a normal random variable with a mean of 1228 and a standard deviation of 120, how many standard deviations are there from 1228 to 1380? a. 11.50 b. 10.233 c. 3.1989 d. 1.267 3. Given that Z is a standard normal variable, which of the following is the value z 0 for which P ( z ≤ z 0 ) = 0 . 242? a. 0.70 b. 0.65 c. –0.65 d. –0.70 4. The time it takes Jessica to bicycle to school is normally distributed, with a mean time of 15 minutes and a variance of 4 minutes. Jessica has to be at school at 8:00 a.m. Suppose it took her 23 minutes to get to school. What can you reasonably infer or conclude? a. Twenty-three minutes to school is not an unusually long commute time. b. A commuting time of 23 minutes is highly unusual or atypical. c. The distribution of commute times must not be normal with mean 15 minutes and standard devi- ation 2 minutes. d. Both b and c 5. If Z is a standard normal random variable, then the area between z = 0 . 0 and z = 1 . 30 is 0 . 4032, while the area between z = 0 . 0 and z = 1 . 50 is 0 . 4332. What is the area between z = –1 . 30 and z = 1 . 50? a. 0.0300 b. 0.0668 c. 0.0968 1

d. 0.8364 6. The scores of a class are normally distributed with a mean of 82 and a standard deviation of 8. What is the probability that the mean score of a sample of 64 students is at least 80? a. 0.0987 b. 0.4772 c. 0.5987 d. 0.9772 Solution 1. 1. d 2. d 3. d 4. d 5. d 6. d See the following graphs Exercise 2. A certain type of automobile battery is known to have a lifespan that is normally distributed, with mean 1100 days and standard deviation 80 days. For how long should these batteries be guaranteed if the manufacturer wants to replace only 5% of the batteries sold because they “died” before the guarantee expired? Solution 2. Let X denotes the lifespan of the battery, given that X ∼ N ( µ = 1100 , σ = 80), then P ( X < x 0 ) = 0 . 05 = ⇒ P ( z < ( x 0 − 1100) / 80) = 0 . 05 = ⇒ ( x 0 − 1100) / 80 = − 1 . 645 . which implies that x 0 = 968 . 4 days 2

Normal Curve, mean = 100 , SD = 20 Shaded Area = 0.2956 x density 0.000 0.005 0.010 0.015 0.020 105 125 Normal Curve, mean = 100 , SD = 20 Shaded Area = 0.2005 x density 0.000 0.005 0.010 0.015 0.020 83.2 Exercise 4. A normal random variable X has an unknown mean µ and a standard deviation σ = 25. If the probability that X exceeds 7 . 5 is 0 . 8289, find µ . Solution 4. It is given that X is normally distributed with σ = 25 but with unknown mean µ , and that P ( X > 7 . 5) = 0 . 8289. In terms of the standard normal random variable Z , we can write, P ( X > 7 . 5) = P [ Z > (7 . 5 − µ ) / 25] = 0 . 8289 = ⇒ P [ Z ≤ (7 . 5 − µ ) / 25] = 1 − 0 . 8289 = ⇒ P [ Z ≤ (7 . 5 − µ ) / 25] = 0 . 1711 = ⇒ (7 . 5 − µ ) / 25 = − 0 . 95 . This implies µ = 25 × 0 . 95 − 7 . 5 = 31 . 25. Exercise 5. A random variable X is normally distributed with µ = 100 and σ = 20. a. What is the median of this distribution? b. Find P ( X ≥ median). c. Find P ( X ≤ 75) Solution 5. a. Since the normal distribution is symmetric, the mean and the median have the same value, so the median is 100. b. P ( X ≥ median) = P ( X ≥ 100) = P ( Z ≥ 0) = 0 . 50. c. P ( X ≤ 75) = P ( Z ≤ − 1 . 25) = 0 . 50 − 0 . 3944 = 0 . 1056 4

Normal Curve, mean = 100 , SD = 20 Shaded Area = 0.5 x density 0.000 0.005 0.010 0.015 0.020 100 Normal Curve, mean = 100 , SD = 20 Shaded Area = 0.1056 x density 0.000 0.005 0.010 0.015 0.020 75 100 Exercise 6. Let Z denote a standard normal random variable. a. Find P ( Z ≥ 1 . 48) b. Find P ( − 0 . 44 < Z < 2 . 68) c. Determine the value of z 0 which satisfies P ( Z ≤ z 0 ) = 0 . 7995. d. Find P ( Z < − 0 . 87) e. Find P ( − 1 . 66 < Z < − 0 . 48) f. Find z 0 such that P ( − z 0 < Z < z 0 ) = 0 . 901 g. Find z 0 such that P ( Z ≤ z 0 ) = 0 . 0375 h. Find z 0 such that P ( − z 0 < Z < z 0 ) = 0 . 7698 Solution 6. Let Z denote a standard normal random variable. a. P ( Z ≥ 1 . 48) = 0 . 0694 b. P ( − 0 . 44 < Z < 2 . 68) = 0 . 6663 c. z 0 = 0 . 84 d. P ( Z < − 0 . 87) = 0 . 1922 e. P ( − 1 . 66 < Z < − 0 . 48) = 0 . 2671 f. z 0 = 1 . 65 g. z 0 = –1 . 78 5

Normal Curve, mean = 104 , SD = 16 Shaded Area = 0.05 x density 0.000 0.005 0.010 0.015 0.020 0.025 104 130.32 Exercise 8. Suppose Z has a standard normal distribution. Then between which two z-values (symmetrically dis- tributed around the mean) would 59 . 9% of the possible z-values occur? Solution 8. 59 . 9% of the possible z-values will occur between − 0 . 8398 and 0 . 8398 Normal Curve, mean = 0 , SD = 1 Shaded Area = 0.599 x density 0.0 0.1 0.2 0.3 0.4 -0.8398 0.8398 7

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 -3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 -3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 -3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 -3.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 -3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010 -2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 -2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 -2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 -2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 -2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 -2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 -2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084 -2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 -2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 -2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 -1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 -1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 -1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 -1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 -1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 -1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 -1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 -1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 -1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 -1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 -0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611 -0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 -0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 -0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 -0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 -0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 -0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 -0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 -0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 Standard Normal Cumulative Probability Table Cumulative probabilities for NEGATIVE z-values are shown in the following table: