1. ## [SOLVED] Normal Distribution

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

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

Attempt:

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

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

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

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

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

$\displaystyle P(X = 15)$

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

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

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

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

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

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

Attempt:

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

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

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

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

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

$\displaystyle P(X = 15)$

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

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

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

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).