Section 7.1 (p.382)
1. 
has a normal distribution with mean q and variance 4/n. Therefore,
4/n £ 0.1 ̃ n ³
40.
3. 
has a standard normal distribution. Therefore,
From Table on p.689 we
know that 
³ 1.96. Therefore n ³ 1537.
4. 
has a binomial distribution with parameters n and p = 0.2
To find the smallest n
satisfying the above inequality, we use “try and error” for some n. Then we can
show that the smallest n is 10. (Try it yourself using the table on p.682. You
will see that n = 7, 8, 9 don’t work and when n = 10, Pr(1 £ SXi
£ 3) = 0.7717 ³
0.75)
5. 
has mean p = 0.2 and variance p(1-p) /n = 0.16/n
will have
approximately a standard normal distribution according to CLT.
Therefore, 
³ 1.96, n ³
62.
7. 
Note that p(1-p) £ ¼ (why? Take a derivative with respect to p).
Therefore for p(1-p)/n £ 0.01 holds for all possible p, n ³ 0.25/0.01 = 25.