In families with 6 children, let X= the number of boys. For simplicity, assume that births are independent and boys and girls are equally likely. a. Graph the probability distribution of X. b. Calculate the mean and standard deviation, and show them on the graph. c. Of all families with 6 children, what proportion have: i. Exactly an even split between the sexes (3-3)? ii. Nearly an even split (3-3, or 4-2)? iii. 3 or more boys? please, write the answer and explain how to calculate???
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if sex of children is equally likely the. Pr of a boy (p) = 0.5 and Pr of a girl(q) = 0.5
Binomial distribution
[nCx]*[(p)^x]*[(q)^(n - x)
[6!/x!(6-x)!]*[(0.5)^x]*[0.5^(6 - x)]
To get the points on your graph calculate the above formula for x = 0 to 6
remember that by definition 0! = 1
for i) x = 3
ii) x = 3 for 3-3, and x = 4 for 4-3
iii) For three or more boys sum the probabilities for x = 3,4,5,6
mean = n*p =6*(0/5)
standard deviation = Square root(n*p*q) = Square root(6*05*0.5) = Square root(1.5) = 1.225
Binomial distribution
[nCx]*[(p)^x]*[(q)^(n - x)
[6!/x!(6-x)!]*[(0.5)^x]*[0.5^(6 - x)]
To get the points on your graph calculate the above formula for x = 0 to 6
remember that by definition 0! = 1
for i) x = 3
ii) x = 3 for 3-3, and x = 4 for 4-3
iii) For three or more boys sum the probabilities for x = 3,4,5,6
mean = n*p =6*(0/5)
standard deviation = Square root(n*p*q) = Square root(6*05*0.5) = Square root(1.5) = 1.225