Question

for a normal variable $xsim n(mu = 55.6,sigma = 10.8)$, find the probability $p(x > 45.8)$: $p(x > 45.8)=\text{blank}$ (round the answer to 4 decimal places)

Explanation:

Step1: Standardize the value

We use the formula $Z=\frac{X - \mu}{\sigma}$. Here, $X = 45.8$, $\mu=55.6$ and $\sigma = 10.8$. So $Z=\frac{45.8 - 55.6}{10.8}=\frac{-9.8}{10.8}\approx - 0.91$.

Step2: Find the probability

We want $P(X>45.8)$, which is equivalent to $P(Z>- 0.91)$ in the standard - normal distribution. Since $P(Z > z)=1 - P(Z\leq z)$, and from the standard - normal table $P(Z\leq - 0.91)=0.1814$. So $P(Z>-0.91)=1 - 0.1814 = 0.8186$.

Answer:

$0.8186$