Answer
$P(125 \leq x \leq 135)$ $\approx$ 0.5536
Work Step by Step
i) Verify that we can use the normal approximation to the binomial
$np(1-p) \geq 10$
= $135(0.90)(0.10) \geq$ 10
= $13.5 \geq 10$
Yes, we can use the normal distribution to approximate the binomial
ii) Find mean and standard deviation of the data
$\mu = np = 150 \times 0.90 = 135$
$\sigma = \sqrt {np(1-p)} = \sqrt{13.5}$
iii) Want to find $P(125 \leq x \leq 135)$
Apply the continuity correction: $P(125 \leq x \leq 135)$ = $P(124.5 \leq x \leq 135.5)$
iv) Find $P(124.5 \leq x \leq 135.5)$
Convert 124.5 to a z score
z = $\frac{124.5-135}{\sqrt{13.5}}= -2.86$
Convert 135.5 to a z score
z = $\frac{135.5-135}{\sqrt{13.5}}= 0.14$
Therefore, $P(124.5 \leq x \leq 135.5)= P(-2.86 < z < 0.14) = P(z < 0.14) - P(-2.86) = 0.5557 - 0.0021 = 0.5536$
