a normal distribution has mean $mu$ and standard deviation $sigma$. find $p(xgeqmu - sigma)$ for a randomly…

Explanation:

Step1: Standardize the value

We use the z - score formula $z=\frac{x - \mu}{\sigma}$. Given $x=\mu-\sigma$, then $z=\frac{(\mu - \sigma)-\mu}{\sigma}=\frac{\mu - \sigma-\mu}{\sigma}=- 1$. So, $P(x\geq\mu - \sigma)=P(z\geq - 1)$.

Step2: Use the properties of the standard normal distribution

The total area under the standard - normal curve is 1. We know that the standard normal distribution is symmetric about $z = 0$. Also, $P(z\geq - 1)=1 - P(z\lt - 1)$. Looking up the value of $P(z\lt - 1)$ in the standard - normal table, we find that $P(z\lt - 1)=0.1587$. Then $P(z\geq - 1)=1 - 0.1587 = 0.8413$.

Answer:

$0.8413$