STAT 210TEST 4 SOLUTIONS – Version A1. 0 (zero) – The probability that a normal variable is exactly equal to any value is zero.2. B, C, and D – The Z distribution and all t-distributions all have a bell shape, a mean of zero, and nounusual features. The only difference between the Z and all t-distributions is that they have differentstandard deviations. 3. A – A sampling distribution is the distribution of all values taken by a statistic in a large number ofsimple random samples of the same size from the same population.4. 0 (zero) – The mean of the standard normal (Z) distribution is zero.5. Top 1.32% implies find x such that P(X > x) = 0.0132.1 – 0.0132 = 0.9868, look up p = .9868 in the body of the table, find z = 2.22So x = + z = 52.8 + 2.22(10.3) = 52.8 + 22.866 = 75.666 years oldOn the calculator: invNorm(0.9868, 52.8, 10.3) = 75.669 years old6. P(X < 24) = P(Z<24−52.810.3) = P(Z < –2.80) = 0.0026On the calculator: normalcdf(–1E99, 24, 52.8, 10.3) = 0.00267. P(X > 46) = P(Z>46−52.810.3) = P(Z > –0.66) = 1 – P(Z < –0.66) = 1 – 0.2546 = 0.7454On the calculator: normalcdf(46, 1E99, 52.8, 10.3) = 0.7454 8. P(28 < X < 37) = P(28−52.810.3<Z<37−52.810.3) =P(–2.41 < Z < –1.53) = P(Z < –1.53) – P(Z < –2.41) = 0.0630 – 0.0080 = 0.0550On the calculator: normalcdf(28, 37, 52.8, 10.3) = 0.0545 9. Bottom 11.12% implies find x such that P(X < x) = 0.1112Look up p = .1112 in the body of the table, find z = –1.22So x = + z = 52.8 + (–1.22)(10.3) = 52.8 – 12.566 = 40.234 years oldOn the calculator: invNorm(0.1112, 52.8, 10.3) = 40.232 years oldSTAT 210TEST 4 SOLUTIONS – Version B1. A, C, and D – The Z distribution and all t-distributions all have a bell shape, a mean of zero, and nounusual features. The only difference between the Z and all t-distributions is that they have differentstandard deviations. 2. C – A sampling distribution is the distribution of all values taken by a statistic in a large number ofsimple random samples of the same size from the same population.3. 1 (one) – The standard deviation of the standard normal (Z) distribution is one.4. 0 (zero) – The probability that a normal variable is exactly equal to any value is zero.5. P(40 < X < 57) = P(40−49.311.8<Z <57−49.311.8) =P(–0.79 < Z < 0.65) = P(Z < 0.65) – P(Z < –0.79) = 0.7422 – 0.2148 = 0.5274On the calculator: normalcdf(40, 57, 49.3, 11.8) = 0.5277 6. Top 1.46% implies find x such that P(X > x) = 0.0146.1 – 0.0146 = 0.9854, look up p = .9854 in the body of the table, find z = 2.18So x = + z = 49.3 + 2.18(11.8) = 49.3 + 25.724 = 75.024 years old On the calculator: invNorm(0.9854, 49.3, 11.8) = 75.033 years old7. Bottom 18.14% implies find x such that P(X < x) = 0.1814Look up p = .1814 in the body of the table, find z = –0.91So x = + z = 49.3 + (–0.91)(11.8) = 49.3 – 10.738 = 38.562 years oldOn the calculator: invNorm(0.1814, 49.3, 11.8) = 38.561 years old8. P(X < 31) = P(Z<31−49.311.8) = P(Z < –1.55) = 0.0606On the calculator: normalcdf(–1E99, 31, 49.3, 11.8) = 0.06059. P(X > 62) = P(Z>62−49.311.8) = P(Z > 1.08) = 1 – P(Z < 1.08) = 1 – 0.8599 = 0.1401On the calculator: normalcdf(62, 1E99, 49.3, 11.8) = 0.1409STAT 210TEST 4 SOLUTIONS – Version C1. C – A sampling distribution is the distribution of all values taken by a statistic in a large number of simple random samples of the same size from the same population.2. 0 (zero) – The mean of the standard normal (Z) distribution is zero.3. 0 (zero) – The probability that a normal variable is exactly equal to any value is zero.4. A, B, and D – The Z distribution and all t-distributions all have a bell shape, a mean of zero, and nounusual features. The only difference between the Z and all t-distributions is that they have differentstandard deviations. 5. P(X > 27) = P(Z>27−47.812.6) = P(Z > –1.65) = 1 – P(Z < –1.65) = 1 – 0.0495 = 0.9505On the calculator: normalcdf(27, 1E99, 47.8, 12.6) = 0.9506 6. P(52 < X < 64) = P(52−47.812.6<Z<64−47.812.6) =P(0.33 < Z < 1.29) = P(Z < 1.29) – P(Z < 0.33) = 0.9015 – 0.6293 = 0.2722On the calculator: normalcdf(52, 64, 47.8, 12.6) = 0.2702 7. Top 2.62% implies find x such that P(X > x) = 0.0262.1 – 0.0262 = 0.9738, look up p = .9738 in the body of the table, find z = 1.94So x = + z = 47.8 + 1.94(12.6) = 47.8 + 24.444 = 72.244 years old On the calculator: invNorm(0.9738, 47.8, 12.6) = 72.242 years old8. Bottom 13.35% implies find x such that P(X < x) = 0.1335Look up p = .1335 in the body of the table, find z = –1.11So x = + z = 47.8 + (–1.11)(12.6) = 47.8 – 13.986 = 33.814 years oldOn the calculator: invNorm(0.1335, 47.8, 12.6) = 33.814 years old9. P(X < 74) = P(Z<74−47.812.6) = P(Z < 2.08) = 0.9812On the calculator: normalcdf(–1E99, 74, 47.8, 12.6) =
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