Question

which of the values of p and n below would make it not acceptable to use a normal approximation for the binomial distribution? p = 0.46, n = 80 p = 0.46, n = 80 p = 0.48, n = 90 p = 0.48, n = 90 p = 0.89, n = 92 p = 0.89, n = 92 p = 0.91, n = 94 p = 0.91, n = 94 p = 0.95, n = 200

Answer 1

Hi there!

$\boxed{n = 90, p = .89 \text{ ,} n = 94, p = .91}$

For values of 'p' and 'n' to be appropriate for a normal approximation:

$\large\boxed{np \geq 10 \text{ and } n(1-p) \geq 10}$

BOTH conditions must be satisfied.

We can go through each choice:

Choice 1:

80(0.46) = 36.8 ≥ 10 (WORKS)

80(1 - .46) = 43.2 ≥ 10 (WORKS)

Choice 2:

90(0.48) = 43.2 ≥ 10 (WORKS)

90(1 - .48) = 46.8 ≥ 10 (WORKS)

Choice 3:

90(.89) = 80.1 ≥ 10 (WORKS)

90(1 - .89) = 9.9 ≤ 10 (DOES NOT WORK)

Choice 4:

92(0.89) = 81.88 ≥ 10 (WORKS)

92(1 - .89) = 10.12 ≥ 10 (WORKS)

Choice 5:

94(0.91) = 85.54 ≥ 10 (WORKS)

94(1 - .91) = 8.46 ≤ 10 (DOES NOT WORK)

Choice 6:

200(.95) = 190 ≥ 10 (WORKS)

200 (1 - .95) = 10 ≥ 10 (WORKS)