Qnt 561 Week 1 Individual Assignment

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QNT/561: Week One Assignment Exercises 80, 82, and 87 (Ch. 3) Exercise 80* a. The times are a population because we are considering the wait times for all of the customers seated on Saturday night. b. To find the mean: µ = ∑ X N µ = 1021 25 µ = 40.84 To find the median: The midpoint value of the population is 39. c. To find the range: Range = Largest Value – Smallest Value Range = 67 – 23 Range = 44 To find the standard deviation: σ = √∑ (X - µ)2 N σ = √∑(X – 40.84) 2 = √5291.36 = √211.65 = 14.55 25 25 Exercise 82* a. To find the mean cost: X = ∑fM = 7,060 = $141.20 N 50 b. To find the standard deviation: s = √∑f (M - X)2 =…show more content…
When X = 29, z = -1.5 When X = 34, z = 1 P(-1.5 < z < 1) = 0.4332 + 0.3413 = 0.7745 = 77.45% c. z = X - µ σ z = 28.7 – 32 = -3.3 = -1.65 2 2 P( z < -1.65) = .0495 = 4.95% d. .5 - .05 = .45 P(z < 1.645) = .05 = 5% z = X - µ σ 1.645 = X – 32 2 3.29 = X – 32 X = 3.29 + 32 = 35.29 hours Exercise 45 z = X - µ σ µ = $1,280 σ = $420 a. When X = $1,500 z = 1500 – 1280 = 220 = .52381 420 420 P(z > .52381) = .300206 = 30.02% b. When X = $1,500, z = .52381 When X = $2,000 z = 2000 – 1280 = 720 = 1.7143 420 420 P(.52381 > z > 1.7143) = .557173 - .300206 = .256967 = 25.7% c. When X = $0 z = 0 – 1280 = -1280 = -3.04762 420 420 P(z < -3.04762) = .001153 = .12% d. .5 - .1 = .4000 P(z < 1.281) = .10 = 10% z = X - µ σ 1.281 = X – 1280 420 538.02 = X – 1280 X = 538.02 + 1280 =

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