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Standard Units and Area Worksheet Answers 1. A normal distribution has 𝜇 = 10 and 𝜎 = 2 a) Find the z – score corresponding to 𝑥 = 12 𝑧= 𝑥 − 𝜇 12 − 10 2 = = = 1.00 𝜎 2 2 b) Find the z – score corresponding to 𝑥 = 4 𝑧= 𝑥 − 𝜇 4 − 10 −6 = = = −3 𝜎 2 2 c) Find the raw score corresponding to 𝑧 = 1.5 𝑥 = 𝑧𝜎 + 𝜇 = 1.5(2) + 10 = 3 + 10 = 13 d) Find the raw score corresponding to 𝑧 = −1.2 𝑥 = 𝑧𝜎 + 𝜇 = −1.2(2) + 10 = −2.4 + 10 = 7.6 2. Use the left tailed z – score table to find each area for each : Area to the left of 𝑧 = −1.26 area = 0.1038 ( from the table ) Area to the left of 𝑧 = 2.54 area = 0.9945 ( from the table ) Area to the left of 𝑧 = 0.55 area = 0.7088 ( from the table ) Area to the left of 𝑧 = −2.71 area = 0.0034 ( from the table ) Area to the left of 𝑧 = 3.02 area = 0.9987 ( from the table ) Area between 𝑧 = −1.45 and 𝑧 = 0.50 Area = (area z = 0.50 – area z = -1.45 ) = 0.6915 – 0.0735 = 0.618 Area between 𝑧 = 1.6 and 𝑧 = 2.75 Area = (area z = 2.75 – area z = 1.6 ) = 0.9970 – 0.9452 = 0.0518 #2 continued : Area between 𝑧 = 1.42 and 𝑧 = 2.17 Area = (area z = 2.17 – area z = 1.42) = 0.9850 – 0.9222 = 0.0628 Area to the right of 𝑧 = −1.05 Area = 1 – (area z = -1.05) = 1 – 0.1469 = 0.8531 Area to the right of 𝑧 = 1.95 Area = 1 – (area z = 1.95) = 1 – 0.9744 = 0.0256 3. The college physical education department offered an advanced first aid course last semester. The scores on the comprehensive final exam were normally distributed, and some of the z – scores are shown below : Robert = 1.10 Jacob = 1.70 Susan = - 2.00 Joel = 0.00 Rachel = - 0.80 Lydia = 1.60 a) Who scored above the mean ? Robert , Jacob , and Lydia b) Who scored below the mean ? Susan , and Rachel c) If the mean score was 𝜇 = 150, with a standard deviation 𝜎 = 20, what was the final exam score for each student ? Robert = 𝑧𝜎 + 𝜇 = 1.10(20) + 150 = 22 + 150 = 172 Joel = 𝑧𝜎 + 𝜇 = 0.00(20) + 150 = 0 + 150 = 150 Jacob = 𝑧𝜎 + 𝜇 = 1.70(20) + 150 = 34 + 150 = 184 Rachel = 𝑧𝜎 + 𝜇 = −0.80(20) + 150 = −16 + 150 = 134 Susan = 𝑧𝜎 + 𝜇 = −2.00(20) + 150 = −40 + 150 = 110 Lydia = 𝑧𝜎 + 𝜇 = 1.60(20) + 150 = 32 + 150 = 182 4. Fawns between 1 and 5 months old in Mesa Verde National Park have a body weight that is approximately normally distributed with mean 𝜇 = 27.2 𝑘𝑔 and standard deviation 𝜎 = 4.3 𝑘𝑔. Let 𝑥 be the weight of a fawn in kilograms. Convert each of the following 𝑥 intervals into 𝑧 intervals . a) 𝑥 < 30 𝑧= 𝑥−𝜇 𝜎 = 30−27.2 4.3 = 2.8 4.3 = 0.65 𝑧 < 0.65 b) 𝑥 > 19 𝑧= 𝑥−𝜇 𝜎 = 19−27.2 4.3 = −8.2 4.3 = −1.91 𝑧 > −1.91 c) 32 < 𝑥 < 35 𝑧= 𝑥−𝜇 𝜎 = 32−27.2 4.3 = 4.8 4.3 = 1.12 𝑧= 𝑥−𝜇 𝜎 = 35−27.2 4.3 = 7.8 4.3 1.12 < 𝑧 < 1.81 Convert each of the following 𝑧 intervals into 𝑥 intervals d) 𝑧 > −2.17 𝑥 = 𝑧𝜎 + 𝜇 = −2.17(4.3) + 27.2 = −9.33 + 27.2 = 17.87 𝑥 > 17.87 e) 𝑧 < 1.28 𝑥 = 𝑧𝜎 + 𝜇 = 1.28(4.3) + 27.2 = 5.50 + 27.2 = 32.7 𝑥 < 32.7 f) −1.99 < 𝑧 < 1.44 𝑥 = 𝑧𝜎 + 𝜇 = −1.99(4.3) + 27.2 = −8.56 + 27.2 = 18.64 𝑥 = 𝑧𝜎 + 𝜇 = 1.44(4.3) + 27.2 = 6.19 + 27.2 = 33.39 18.64 < 𝑥 < 33.39 = 1.81 g) What is the probability that a fawn weighs between 22.5 kg and 33.5 kg ? 𝑧= 𝑧= 𝑥 − 𝜇 22.5 − 27.2 −4.7 = = = −1.09 𝜎 4.3 4.3 𝑥−𝜇 𝜎 = 33.5−27.2 4.3 = 6.3 4.3 = 1.47 Area = ( area between z = -1.09 and z = 1.47) Area = ( area z = 1.47) – (area z = - 1.09) Area = 0.9292 – 0.1379 = 0.7913 5. Let 𝑧 be a random variable with a standard normal distribution. Find the indicated probability : a) 𝑃(𝑧 ≤ −0.13) = 0.4483 b) 𝑃(𝑧 ≤ 3.20) = 0.9993 c) 𝑃(𝑧 ≥ −1.50) = 1 − 0.0668 = 0.9332 d) 𝑃(−1.20 ≤ 𝑧 ≤ 2.64) = 0.9959 − 0.1151 = 0. e) 𝑃(−1.78 ≤ 𝑧 ≤ −1.23) = 0.1093 − 0.0375 = 0.0718 f) 𝑃(0 ≤ 𝑧 ≤ 1.62) = 0.9474 − 0.5000 = 0.4474