Question

for a normal variable $xsim n(mu = 73,sigma = 3.8)$, find the probability $p(x > 71.8)$: $p(x > 71.8)=\text{blank}$ (round the answer to 4 decimal places) question help: video message instructor post to forum submit question

Explanation:

Step1: Standardize the value

We use the formula $Z=\frac{X - \mu}{\sigma}$. Here, $X = 71.8$, $\mu=73$, $\sigma = 3.8$. So $Z=\frac{71.8 - 73}{3.8}=\frac{- 1.2}{3.8}\approx - 0.32$.

Step2: Find the probability

We want $P(X>71.8)$, which is equivalent to $P(Z>-0.32)$ in the standard - normal distribution. Since $P(Z > z)=1 - P(Z\leq z)$, and from the standard - normal table $P(Z\leq - 0.32)=0.3745$. Then $P(Z>-0.32)=1 - 0.3745 = 0.6255$.

Answer:

$0.6255$