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

a normal distribution has mean $mu$ and standard deviation $sigma$. find $p(xleqmu - sigma)$ for a randomly selected $x$-value from the distribution.\n$p(xleqmu - sigma)=square$

a normal distribution has mean $mu$ and standard deviation $sigma$. find $p(xleqmu - sigma)$ for a randomly selected $x$-value from the distribution.\n$p(xleqmu - sigma)=square$

Answer

Explanation:

Step1: Standardize the value

We use the z - score formula $z=\frac{x - \mu}{\sigma}$. Here, $x=\mu-\sigma$. Substituting into the formula, we get $z=\frac{(\mu - \sigma)-\mu}{\sigma}=\frac{- \sigma}{\sigma}=- 1$.

Step2: Use the standard normal table

We want to find $P(x\leq\mu - \sigma)$, which is equivalent to $P(z\leq - 1)$ in the standard - normal distribution (where the standard - normal distribution has mean 0 and standard deviation 1). Looking up the value of $P(z\leq - 1)$ in the standard - normal table (also called the z - table), we know that the total area under the standard - normal curve is 1, and the table gives the area to the left of a z - value. From the standard - normal table, $P(z\leq - 1)=0.1587$.

Answer:

$0.1587$