1. ## [SOLVED] Normal Distribution

Question:
The random variable, $X$, has a binomial distribution with parameters $n = 40$ and $p = 0.3$. Use a suitable approximation, which you should show in valid, to calculate the following probability: $P(X = 15)$

(I should answer this question using Normal Distribution Table and not by integration!)

Attempt:

$P(X = 15)$████ $n = 40$████ $p = 0.3$████ $q = 1 - 0.3 = 0.7$

$np = 40 \times 0.3 = 12$████ $nq = 40 \times 0.7 = 28$

$np > 5$████ $nq > 5$████ So, the equation is VALID

$\mu = np = 12$
$\sigma^2 = npq = 40 \times 0.3 \times 0.7 = 8.4$
$\sigma = \sqrt{8.4}$

$X \sim B(40 , 0.3)$
$X \sim V(np, npq)$
$X \sim N(\mu , \sigma^2)$
$X \sim N(12 , 8.4)$

$P(X = 15)$

$P(14.5 \leq V \leq 15.5)$████ $15 - 0.5 = 14.5$████ $15 + 0.5 = 15.5$

$P\left( \frac{(14.5 - 12)}{\sqrt{8.4}} \leq \frac{V - \mu}{\sigma} \leq \frac{(15.5 - 12)}{\sqrt{8.4}} \right)$

$P( 0.863 \leq Z \leq 1.208 )$
$= \phi(1.208) - \phi(0.863)$
$= (0.8849 + 0.0015) - (0.8023 + 0.0008)$
$= 0.0833$

Answer in textbook is 0.0808, where did I go wrong?

2. Originally Posted by looi76
Question:
The random variable, $X$, has a binomial distribution with parameters $n = 40$ and $p = 0.3$. Use a suitable approximation, which you should show in valid, to calculate the following probability: $P(X = 15)$

(I should answer this question using Normal Distribution Table and not by integration!)

Attempt:

$P(X = 15)$████ $n = 40$████ $p = 0.3$████ $q = 1 - 0.3 = 0.7$

$np = 40 \times 0.3 = 12$████ $nq = 40 \times 0.7 = 28$

$np > 5$████ $nq > 5$████ So, the equation is VALID

$\mu = np = 12$
$\sigma^2 = npq = 40 \times 0.3 \times 0.7 = 8.4$
$\sigma = \sqrt{8.4}$

$X \sim B(40 , 0.3)$
$X \sim V(np, npq)$
$X \sim N(\mu , \sigma^2)$
$X \sim N(12 , 8.4)$

$P(X = 15)$

$P(14.5 \leq V \leq 15.5)$████ $15 - 0.5 = 14.5$████ $15 + 0.5 = 15.5$

$P\left( \frac{(14.5 - 12)}{\sqrt{8.4}} \leq \frac{V - \mu}{\sigma} \leq \frac{(15.5 - 12)}{\sqrt{8.4}} \right)$

$P( 0.863 \leq Z \leq 1.208 )$
$= \phi(1.208) - \phi(0.863)$
$= (0.8849 + 0.0015) - ({\color{red}0.8023} + 0.0008)$
$= 0.0833$

Answer in textbook is 0.0808, where did I go wrong?
Your error is in red. It should be 0.8051 (you looked at the '5 column' by mistake).