Answer
$P(X = 60) = \frac{75!}{60!(75-60)!} \times 0.75^{60} \times (1 - 0.75)^{(75-60)}= 0.0677$
$P(z < 1.13) - P(z < 0.87) = 0.0630$
Work Step by Step
Here we have: n = 75, p = 0.75, x = 60
Using the binomial probability formula:
$P(X = 60) = \frac{75!}{60!(75-60)!} \times 0.75^{60} \times (1 - 0.75)^{(75-60)}$
= 0.0677
Check whether the normal distribution can be used as an approximation for the binomial distribution:
$np(1-p) = 75 x 0.75 (1 - 0.75) = 14.1 \gt 10$
Hence, the normal distribution can be used.
$μ_{x} = np = 75 \times 0.75 = 56.25$
$σ_{x} = \sqrt {np(1-p)} = \sqrt {56.25 (0.25)} = 3.75$
$z = \frac{x - μ_{x}}{σ_{x}} = \frac{59.5 - 56.25}{3.75} =0.87$
$z = \frac{x - μ_{x}}{σ_{x}} = \frac{60.5 - 56.25}{3.75} = 1.13$
$P(X = 60) = P(59.5 < X < 60.5) = P(0.87 < z < 1.13)$
$P(z < 1.13) - P(z < 0.87) = 0.8708 - 0.8078 = 0.0630$
