Answer
$P(x \geq 130)$= 0.9332
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 $P(x \geq 130)$
Apply the continuity correction: $P(x \geq 130)$ = $P(x \geq 129.5)$
iv) Find $P(x \geq 129.5)$
Convert 129.5 to a z score
z = $\frac{129.5-135}{\sqrt{13.5}}= -1.50$
Therefore, $P(x \geq 129.5)$= $P(z > -1.50)$ $= 1- 0.0668$ = 0.9332
