Answer
$P(70 \leq z \leq 90) = 0.9914$
Work Step by Step
Here we have: n = 100, p = 0.80, $70 \leq x \leq 90$
Check whether the normal distribution can be used as an approximation for the binomial distribution:
$np(1-p) = 100 x 0.80 (1 - 0.80) = 16 \gt 10$
Hence, the normal distribution can be used.
$μ_{x} = np = 100 \times 0.80 = 80$
$σ_{x} = \sqrt {np(1-p)} = \sqrt {80(.20)} = 4$
After applying the continuity correction, we have:
$z = \frac{x - μ_{x}}{σ_{x}} = \frac{69.5 - 80}{4} = -2.625$
$z = \frac{x - μ_{x}}{σ_{x}} = \frac{90.5 - 80}{4} = 2.625$
$P(69.5 \leq x \leq 90.5) = P(-2.625 \leq z \leq 2.625)$
$P(z \leq 2.625) - P(z \leq -2.625) = 0.9957 - 0.0043 = 0.9914$
