Exam1 A_F18_Key-1.pdf

# Z score corresponding to lower tail 1 233 ok to use

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z -score corresponding to lower tail 1%: -2.33. OK to use -2.32. In Form B: -2.05 or -2.06 Using the back-calculations, we get X = - 2 . 33 × 15 + 170 = 135 . 05 . (If choosing z = - 2 . 32 , the answer is 135 . 20 . ) For Form B: X = - 2 . 05 × 20 + 165 = 124 . 00 or 123 . 80 for z = - 2 . 06 (c) [6pts.] Ray has the wild idea of plowing 3 acres of his land to build a baseball field. Assuming corn prices are \$3.50 per bushel , how much revenue would he be losing if his farm’s corn yield is in the 80th percentile , assuming every acre has the same yield? For full credit, report the z -score, yield (in bushels per acre), and revenue (in \$). Round your final answer to the nearest dollar. First, calculate the z-score: P ( Z < z ) = 0 . 8 and thus z = 0 . 84 (or 0.85, or 0.845) For Form B: P ( Z < z ) = 0 . 95 and thus z = 1 . 64 (or 1.65, or 1.645) Corresponding yield: X = 0 . 84 × 15 + 170 = 182 . 6 or . 85 × 15 + 170 = 182 . 75 For Form B: X = 1 . 64 × 20 + 165 = 197 . 8 or 1 . 65 × 20 + 165 = 198 . 0 Revenue for all 3 acres: 182 . 6 × 3 × 3 . 50 = \$1917 . 30 or 182 . 75 × 3 × 3 . 50 = \$1918 . 88 , so accept answers between \$ 1917 and \$ 1919. For Form B: 197 . 8 × 4 × 3 . 75 = \$2967 or 182 . 75 × 3 × 3 . 50 = \$2970 , so accept answers between \$ 2967 and \$ 2970. 6. [2pts.] Normal Distribution. Assume you have a normal distribution with mean μ = - 9000 and standard deviation σ = 0 . 01. What is the total area under the curve associated with this distribution? The area under the curve equals 1. The area under the distribution curve of any distribution is 1 (explanation not needed for full credit) -score:

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Exam I Stat 226 Fall 2018 —KEY Page 7 7. Laundry detergent use. The amount of laundry detergent that customers use has a right-skewed distribution with mean of 30 milliliters (mL) and a standard deviation of 25 mL. (a) [3pts.] Suppose we conducted a study in which we measured the laundry detergent use of 4 randomly sampled US adults. The corresponding sampling distribution of the sample mean, i.e. the distribution of all sample means based on all samples of size 4 from all US adults, has the following attributes i. Shape (Choose one): Normal Approximately Normal None of the above ii. Mean: \$30 , B: \$20 Standard Error: \$25 / 4 = \$12 . 5 , B: \$15 / 4 = \$7 . 5 (b) [3pts.] Suppose we conducted a study in which we measured the laundry detergent use of 64 randomly sampled US adults. The corresponding sampling distribution of the sample mean, i.e. the distribution of all sample means based on all samples of size 64 from all US adults, has the following attributes: i. Shape (Choose one): Normal Approximately Normal None of the above ii. Mean: \$ 30, B: \$20 Standard Error: \$25 / 64 = \$3 . 125 , B: \$15 / 64 = \$1 . 875 (c) [4pts.] Find the approximate probability that the mean laundry detergent use for 64 randomly sampled adults is less than 29. (Report answer as a proportion to 4 decimal places.) Show work here: P ( X < 29) = P X - 30 3 . 125 < 29 - 30 3 . 125 = P ( Z < - 0 . 32) = 0 . 3745 B: P ( X < 19) = P X - 20 1 . 875 < 19 - 20 1 . 875 P ( Z < - 0 . 53) = 0 . 2981 (d) [4pts.] What is the value (in mL) that corresponds to the 95th percentile of the sampling distribution of the sample mean of the 64 adults? (Round your answer to 2 decimal places.) Show work here: z = 1 . 65 , x = 30 + 3 . 125 × 1 . 65 = 35 . 16 B: z = 1 . 65 , x = 20 + 1 . 875 × 1 . 65 = 23 . 09 (e) [2pts.] TRUE or FALSE: Greater than 50% US adults use less than 30 mL of laundry detergent. Explain.
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