The first column of this worksheet is your number of sibings +1 to represent the

Question

The first column of this worksheet is your number of sibings +1 to represent the number of children in the families of your respondents. Create a Probability Distribution for number of children including %s. (a class for each value) Calculate the Mean (expected value or expectation) of number of children using the formula ? = ?xP(X = x) Calculate the Standard Deviation of number of children using the formula ? =? (x ? ?) 2 P(X = x). Calculate number of small families (1,2 or 3 children) and number of large families (more than 3 children). Use the binomial probability formula P(X = x) = nCx(p)x (1 ? p) n?x , x = number of respondents that come from a large family calculate the probability of 2 respondent coming from a large family (more than 3 children) during 3 trials. Calculate the Mean of a Binomial Random Variable with parameters n and p (? = np) Calculate the Standard Deviation of a Binomial Random Variable with parameters n and p (? = ? np(1 ? p))

* This bottom portion is the example the prof. sent us, Please help. I don't know what to do.

Explanation / Answer

the mean of number of children is = xP(X = x)=sum of the 4th column of the table which is already provided as 3.43

hence =3.43 [answer]

the standard deviation of the number of children is given as

=sqrt[ (x ) 2 P(X = x)]=sum of 6th coulmn of the table-3.432=13.97-3.432=2.2051 [answer]

total number of families=sum of 2nd column=30

out of which number of small families are=sum of first 3 rows of the second column=3+6+7=16 [answer]

hence the number of large families=30-16=14 [answer]

now we need to Use the binomial probability formula P(X = x) = nCx(p)x (1 p) n-x , x = number of respondents that come from a large family to calculate the probability of 2 respondent coming from a large family (more than 3 children) during 3 trials. hence n=3

now p=probability of respondents coming from a large family=P[X=4]+P[X=5]+...+P[X=10]=0.47

hence the required probability is P[X=2]=3C2(0.47)2(1-0.47)3-2=0.351231 [answer]

the mean of the binomial random variable is np=3*0.47=1.41 [answer]

and the standard deviation is =sqrt[np(1-p)]=sqrt[3*0.47*(1-0.47)]=0.864465 [answer]

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