Answer
$P(x = 130) \approx 0.044$
Work Step by Step
i) First, we verify that we can use the normal approximation to the binomial :
$np(1-p) \geq 10$
= $150(0.90)(0.10) \geq$ 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 $P(x = 130)$
Apply the continuity correction: $P(x = 130)$ = $P(129.5 \leq x \leq 130.5)$
iv) Find $P(129.5 \leq x \leq 130.5)$
Convert 129.5 and 130.5 to z-scores
z = $\frac{129.5-130}{\sqrt 13.5} = -1.50$
z = $\frac{130.5-130}{\sqrt 13.5} = -1.22$
Therefore, $P(129.5 < x < 130.5)$ = $P(-1.50 < z < -1.22)$
$= P(z < -1.22) - P(z < -1.50)$
$= 0.1112 - 0.0668$
$= 0.0444$
