HW5-solutions

# HW5-solutions - ScottPowers([email protected]

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STAT 22000 Fall 2011 Homework 3 Solution Key Scott Powers ([email protected]) 4.92 Mean of average loss for 10 policies: \$300 Standard deviation for 10 policies: \$126.49 Mean of average loss for 12 policies: \$300 Standard deviation for 12 policies: \$115.47 4.129 (a) If either parent were type aa, then that parent would be albino, but we know this not to be the case. If either parent were type AA, then the brother would have an A allele from that parent and would thus not be albino, which we also know not to be the case. So both parents have type Aa. (b) A child of Beth’s parents could have any of the three types. The probability of type aa is a quarter; the probability of type Aa is a half; and the probability of type AA is a quarter. (c) Given that she is not albino, Beth has type Aa with probability two in three and type AA with probability one in three. 4.130 (a) The probability that a child of Beth and Bob is non‐albino given that Beth has type Aa is one half. Given that Beth has type AA, the probability of a non‐albino child is one.

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HW5-solutions - ScottPowers([email protected]

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