Answer :
Step-by-step explanation:
we have p=proportion of vegetarian students=0.08
a. FALSE.
for normal approximations we must have np>=5 and n(1-p)>=5
here n=60 so np=60*0.08=4.8 which is less than 5.
hence normal approximation is not applicable.
b. TRUE.
for binomial proportion , the distribution is right skewed if 1-2p>0 or, 0.5>p
here, p=0.08 so 0.5>p is satisfied. hence it is rightly skewed.
c. FALSE.
here, n=125 which is quite large. hence normal approximation is applicable.
let phat be the sample proportion
then phat approximately follows a normal distribution with E[phat]=p=0.08
and variance=V[phat]=p(1-p)/n=0.08(1-0.08)/125=0.0005888
hence standard deviation=sqrt(0.0005888)=0.0242652
hence effective range of phat is [0.08-3*0.0242652,0.08+3*0.0242652]=[0.0072044,0.1527956]
12%=0.12 lies in this effective range.
hence it is not unusual.
In this exercise we have to classify whether the statements are true or false:
a) False
b) True
c) False
d) False
e) False
We have: [tex]p=proportion of vegetarian students=0.08[/tex]
a) FALSE.
For normal approximations we must have:
[tex]np\geq 5 \\n(1-p)\geq 5[/tex]
Where [tex]n=60[/tex] so [tex]np=60*0.08=4.8[/tex] which is less than 5. Hence normal approximation is not applicable.
b)TRUE.
For binomial proportion , the distribution is right skewed if :
[tex]1-2p>0 \\ 0.5>p[/tex]
Where, p=0.08 so 0.5>p is satisfied. hence it is rightly skewed.
c) FALSE.
Where, n=125 which is quite large. hence normal approximation is applicable. Let phat be the sample proportion then phat approximately follows a normal distribution with:
[tex]E[phat]=p=0.08[/tex]
and [tex]variance=V[phat]=p(1-p)/n=0.08(1-0.08)/125=0.0005888[/tex]
Hence standard deviation [tex]=sqrt(0.0005888)=0.0242652[/tex]
hence effective range of phat is:
[tex][0.08-3*0.0242652,0.08+3*0.0242652]=[0.0072044,0.1527956][/tex]
12%=0.12 lies in this effective range. Hence it is not unusual.
d) FALSE.
e) FALSE.
See more about proportion at : brainly.com/question/8704765