Answer
$P(X = 18) = \frac{80!}{18!(80-18)!} \times 0.15^{18} \times (1 - 0.15)^{(80-18)}= 0.2208$
$P(z < 2.04) - P(z < 1.72) = 0.0220$
Work Step by Step
Here we have: n = 80, p = 0.15, x = 18
Using the binomial probability formula:
$P(X = 18) = \frac{80!}{18!(80-18)!} \times 0.15^{18} \times (1 - 0.15)^{(80-18)}= 0.2208$
Check whether the normal distribution can be used as an approximation for the binomial distribution:
$np(1-p) = 80 x 0.15 (1 - 0.15) = 10.2\ge 10$
Hence, the normal distribution can be used.
$μ_{x} = np = 80 \times 0.15 = 12$
$σ_{x} = \sqrt {np(1-p)} = \sqrt {12 (0.85)} = 3.19$
$z = \frac{x - μ_{x}}{σ_{x}} = \frac{17.5 - 12}{3.19} = 1.72$
$z = \frac{x - μ_{x}}{σ_{x}} = \frac{18.5 - 24}{3.19} = 2.04$
$P(X = 18) = P(17.5 < X < 18.5) = P(1.72 < z < 2.04)$
$P(z < 2.04) - P(z < 1.72) = 0.9739 - 0.9573 = 0.0220$
