Suppose that 8% of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning. a. The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since n >= 30.
b. The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed.
c. A random sample of 125 college students where 12% are vegetarians would be considered unusual.
d. A random sample of 250 college students where 12% are vegetarians would be considered unusual.
e. The standard error would be reduced by one-half if we increased the sample size from 125 to 250.

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.

£56

step-by-step explanation:

u divide 24 by 3 and u get 8 so if u do 8 time 7 or 8+8+8+8+8+8+8 u get 56 and 56 divided by 7 is 8.

the answer would be 484

step-by-step explanation:

just multiply 0.484 times 1,000 (formula)