Section 6.2 Recitation Problems

# Section 6.2 Recitation Problems - Solutions for Recitation...

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Solutions for Recitation Problems from Section 6.2 Section 6.2 for the normal distribution: Problems 18, 20, 23 18. Let x be the continuous r.v. of the price per share for companies on the Given: μ = \$30.00 σ = \$8.20 x is normal a. Want Prob ( x ≥ \$40) At x = 40, 40 30 1.22 8.2 z - = = By equivalence, Prob ( x ≥ \$40) = Prob ( z ≥ 1.22) From z table, down the leftmost “z” column to “1.2,” across the “1.2” row until under the column “.02,” the table entry for “1.22” is 0.8888. So, Prob ( z 1.22) = 0.8888 But want Prob ( z ≥ 1.22) The area under the entire normal is 1. Also, Prob ( z ≥ 1.22) and Prob ( z ≤ 1.22) are “contiguous” areas, meaning that they adjoin each other without overlap or gap. Then Prob ( z ≤ 1.22) + Prob ( z ≥ 1.22) = 1 Then Prob ( z ≥ 1.22) = 1 – Prob ( z ≤ 1.22) = 1 – 0.8888 = 0.1112 Returning to the original r.v., Section 6.2 recitation problem solutions 1 © Harvey Singer 2009

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Prob ( x ≥ 40) = 0.1112 b. Want Prob ( x ≤ \$20) At x = 20, 20 30 1.22 8.2 z - = = - By equivalence, Prob ( x ≤ \$20) = Prob ( z ≤ -1.22) From negative z table, down the leftmost “z” column to “-1.2,”
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Section 6.2 Recitation Problems - Solutions for Recitation...

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