Answer
$P(99.5 \lt X \leq 100.5) = P(2.38 < z \le 2.5) = 0.0025$
Work Step by Step
Here we have: n = 500, p = 0.16, x = 100
Check whether the normal distribution can be used as an approximation for the binomial distribution:
np(1-p) = 500 x 0.16 (1 - 0.16) = 67.2 > 10
Hence, the normal distribution can be used.
$μ_{x} = np = 500 \times 0.16 = 80$
$σ_{x} = \sqrt {np(1-p)} = \sqrt {80 (0.84)} = 8.20$
$z = \frac{x - μ_{x}}{σ_{x}} = \frac{99.5 - 80}{8.20} = 2.38$
$z = \frac{x - μ_{x}}{σ_{x}} = \frac{100.5 - 80}{8.20} = 2.5$
$P(99.5 \lt X \leq 100.5) = P(2.38 < z \le 2.5) = 0.0025$
