It has 2 solutions i believe
Answer:
0.25
Step-by-step explanation:
We have a total of ten student, and three students are randomly selected (without replacement) to participate in a survey. So, the total number of subsets of size 3 is given by 10C3=120.
On the other hand A=Exactly 1 of the three selected is a freshman. We have that three students are freshman in the classroom, we can form 3C1 different subsets of size 1 with the three freshman; besides B=Exactly 2 of the three selected are juniors, and five are juniors in the classroom. We can form 5C2 different subsets of size 2 with the five juniors. By the multiplication rule the number of different subsets of size 3 with exactly 1 freshman and 2 juniors is given by
(3C1)(5C2)=(3)(10)=30 and
Pr(A∩B)=30/120=0.25
Red : Cream : Blue
2 : 3 : 1
2x2 : 3x2 : 1x2
4 : 6 : 2
You need 6 parts of Cream and 2 parts of Blue
Answer:
C. Decreases the margin of error and hence increases the precision
Step-by-step explanation:
If we select a sample by Simple Random Sampling in a population of “infinite” size (a population so large that we do not know its size exactly), then the margin of error is given by
where
<em>Z = The Z-score corresponding to the confidence level
</em>
<em>S = The estimated standard deviation of the population
</em>
<em>n = the size of the sample.
</em>
As we can see, since n is in the denominator of the fraction and the numerator is kept constant, the larger the sample size the smaller the margin of error, so the correct choice is:
C. Decreases the margin of error and hence increases the precision
