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5.2.q10

# 5.2.q10 - Student Grady Simonton Course Mathll9 Elementary...

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Unformatted text preview: Student: Grady Simonton Course: Mathll9: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, lle Time: 4:23 PM Let the random variable x represent the number of girls in a family with three children. The probability of a child being a girl is 0.4. Construct a table describing the probability distribution, then ﬁnd the mean and standard deviation. Is it unusual for a family of three children to consist of three girls? To ﬁnd the probability of a family having 241 girls out of 3 children, ﬁrst determine and list out the possible outcomes of 3 children. Then use the rule for complementary events, the multiplication rule, and the addition rule. There is 1 possible way a family can have 0 girls out of 3 children: {boy, boy, boy}. There are 3 possible ways a family can have 1 girl out of 3 children: {girl, boy, boy}, {boy, girl, boy}, and {boy, boy, girl}. There are 3 possible ways a family can have 2 girls out of 3 children: { girl, girl, boy}, {girl, boy, girl}, and {boy, girl, girl}. There is 1 possible way a family can have 3 girls out of 3 children: { girl, girl, girl}. The table below displays the possible outcomes of 3 children. x Possible Outcomes {boy, boy, boy} {girL boy. boy}, {boy, girl, boy}, {boy, boy, : irl {girL gir1,boy}, {girL boy, girl}, {boy, girl, girl} Now determine the probabilities of 0, l, 2, and 3 girls out of 3 children. The probability of a child being a girl is 0.4. To find the probability of a child being a boy, use the rule for complementary events. P(boy) = l - P(girl) = 1 - 0.4 = 0.6 Use the multiplication rule and the addition rule to determine the probability of a family having 0, l, 2, and 3 girls out of 3 children. Multiply the probability of each child for each outcome, and add the products of each outcome. For 0 girls out of 3 children, multiply the probability of each child for the outcome {boy, boy, boy}. 0.6 - 0.6 - 0.6 = 0.216 For 1 girl out of 3 children, multiply the probability of each child for each outcome { girl, boy, boy}, {boy, girl, boy}, and {boy, boy, girl}, and add the products of each outcome. Page 1 Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 492.39 Instructor: Shawn Parvini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, 11e Time: 4:23 PM (0.4 - 0.6 - 0.6) +(0.6 - 0.4 - 0.6) +(0.6 - 0.6 - 0.4) =0.432 For 2 girls out of 3 children, multiply the probability of each child for each outcome { girl, girl, boy}, { girl, boy, girl}, and {boy, girl, girl}, and add the products of each outcome. (0.4 - 0.4 - 0.6) +(0.4 - 0.6 - 0.4) +(0.6 - 0.4 - 0.4) =0.288 For 3 girls out of 3 children, multiply the probability of each child for the outcome { girl, girl, girl}. 0.4 ' 0.4 ' 0.4 =0.064 The table below displays the probability distribution of a family having it girls out of 3 children. Find the mean of the random variable using the formula below. Enron x0‘1’(x0)+x1 -P(x1)+x2-P(x2)+x3 -P(x3) 0 ' 0216+] ' 0.432+2 - 0288+ 3 ' 0.064 1.2 it Find the standard deviation of the random variable using the formula below. °=JZBLH9PM = 0x5 ' P(Xo) + Xi' P(X1)+ X3 ' P(XZ) + X? ' 130(3) ‘ 112 =1.102-0.216+12-0.432+22-0.2ss+32-0.064—1.22 = 0.85 (rounded to the nearest hundredth) It would be unusual for a family with three children to consist of three girls if P(3) is less than or equal to 0.05. Page 2. ...
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