Question

Need help with part (C).

7.70 In this section, we provided the rule of thumb that the normal approximation to the binomial distribution is adequate if p +3 pg/n lies in the interval (0, 1)—that is, if 0 < p – 31pq/n and p+31pq/n < 1. a Show that p +37pqn<l if and only if n > 9(p/9). b Show that 0 <p – 31pq/n if and only if n > 9(q/p). C Combine the results from parts (a) and (b) to obtain that the normal approximation to the binomial is adequate if n>9(%) and n>90%). or, equivalently, n>91 larger of p and a smaller of p and 9)

show the equivalence using cases. Case 1: p>=q. Case 2:q>p. Thank you.

Answer 1

Solution for Part (C)

Back-up Theory

If x > y and x > z, then x > max(y, z) .................................................................................. (1a)

If x < y and x < z, then x < min(y, z) .................................................................................. (1b)

Now to work out the solution,

Case 1:

p ≥ q

=> p/q ≥ 1 and q/p ≤ 1 ...........................................................................................................(2)

=> 9(p/q) ≥ 9 and 9(q/p) ≤ 9

=> 9(p/q) ≥ 9(q/p)

Hence, vide (1a), n > 9(p/q)

Since by assumption, p ≥ q, p is larger of p and q and q is smaller of p and q

Thus, n > 9(larger of p and q)/(smaller of p and q) Answer 1

Case 2:

p < q

=> p/q < 1 and q/p > 1 ..................................................................................................................(3)

=> 9(p/q) < 9 and 9(q/p) > 9

=> 9(q/p) > 9(p/q)

Hence, vide (1a), n > 9(q/p)

Since by assumption, p < q, p is smaller of p and q and q is larger of p and q

Thus, n > 9(larger of p and q)/(smaller of p and q) Answer 2

DONE