Part a is correct.
Part b can be done a couple ways.
1) Use binomial distribution. p = 0.2, n = 144
P(x>3) = 1 - P(x=0) -p(x=1) -p(x=2) -p(x=3)
where
[tex]P(x=k) = (nCk)p^k (1-p)^{n-k}[/tex]
2) Assume a normal distribution with mean = 28.2, stdev = 4.8
[tex]z = \frac{4 - 28.2}{4.8} = -5.04[/tex]
Look up z-value in normal table to find probability.
Hope that helps.
