# 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 =
When X = 29, z = -1.5
When X = 34, z = 1
P(-1.5 &lt; z &lt; 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 &lt; -1.65) = .0495 = 4.95%
d. .5 - .05 = .45
P(z &lt; 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 &gt; .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 &gt; z &gt; 1.7143) = .557173 - .300206 = .256967 = 25.7%
c. When X = \$0 z = 0 – 1280 = -1280 = -3.04762 420 420 P(z &lt; -3.04762) = .001153 = .12%
d. .5 - .1 = .4000
P(z &lt; 1.281) = .10 = 10% z = X - µ σ 1.281 = X – 1280 420 538.02 = X – 1280 X = 538.02 + 1280 =