Problem 2 (12 points, 2 point for each part) a. In California, 39% of people ove
ID: 3207887 • Letter: P
Question
Problem 2 (12 points, 2 point for each part) a. In California, 39% of people over the age of 5 speak alanguage other than English. Suppose we take a sample of 45 people overthe age of 5 and determine the languages they speak. i. Determine the sampling distribution of the sample proportion Mean Standard Error: Shape: Reason for Shape: ii. Suppose we want to calculate the probability that more than half of the people in the sample spoke a language other than English. If possible, calculate this probability. If not, briefly explain why it cannot be calculated.Explanation / Answer
12 a) here n=45, p=0.39
(i) p=proporiton of people who spoke other than english
mean=np=45*0.39=17.55,
Standard error=sqrt(np(1-p))=sqrt(45(0.39*(1-0.39))=3.27
shape:symmetrical
reason:sample size is more than 30 and np(1-p)>10
(ii) here x=n/2=45/2=22.5 and z=(x-mean)/SE(p)=(22.5-17.55)/3.27=1.51
P(x>more than half)=P(x>22.5)=P(z>1.51)=1-P(z<1.51)=1-0.935=0.065
12 b) here n=83, p=0.93
(i) p=diagonised and will die=0.93
mean=np=83*0.93=77.19,
Standard error=sqrt(np(1-p))=sqrt(83(0.93*(1-0.93))=2.32
shape:symmetrical
reason:sample size is more than 30 and np>10
(ii) for p=0.9 then x=np=0.9*0.83=74.7 and z=(74.7-77.19)/2.32=-1.07
P(x<74.7)=P(z<-1.07)=0.14
12 c) here n=80, p=0.68
(i) p= pitches were strike=0.68
mean=np=80*0.68=54.4,
Standard error=sqrt(np(1-p))=sqrt(80*(0.68*(1-0.68)))=4.17
shape:symmetrical
reason:sample size is more than 30 and np>10
(ii) for p=0.75 then x=np=0.75*0.80=60and z=(60-54.4)/4.17=1.34
P(x>three four quarter)=P(x>60)=P(z>1.34)=1-P(z<1.34)=1-0.91=0.09