418NormalDistribution_ExtraPractice

# 418NormalDistribution_ExtraPractice - The Normal...

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1 The Normal Distribution Example The average new car costs \$23,000 with a standard deviation (SD)=\$3,500. What proportion of all cars cost \$32,000 or more? What proportion of all cars cost \$16,000 or more? What is the percentile rank for a car that costs \$30,000? What is the percenitle rank for a car that sold for \$12,000? What proportion of cars were sold for an amount between \$12,000 and \$30,000? For what price would a car at the 16th percentile have sold? Given: μ =23,000 σ =3,500 Question: What proportion of cars cost \$32,000 or more? z 1 =32000-23000 3500 = 9000/3500 = 2.57 23000 3500 0 2.57 1 32000

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2 0.00507 0.51% Given: μ =23,000 σ =3,500 Question: What proportion of cars cost \$32,000 or more? z 1 =32000-23000 3500 = 9000/3500 = 2.57 23000 3500 32000 0 2.57 1 Given: μ =23,000 σ =3,500 Question: What proportion of cars cost \$16,000 or more? z 1 =16000-23000 3500 = -7000/3500 = -2 23000 3500 16000 0 1 -2
3 Given: μ =23,000 σ =3,500 Question: What proportion of cars cost \$16,000 or more? z

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418NormalDistribution_ExtraPractice - The Normal...

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