ssume the random variable $x$ is normally distributed with mean $\mu=89$ and standard deviation $\sigma=5$. Find the indicated probability.
\[
P(x< 86)
\]
\[
P(x< 86)=\square \text { (Round to four decimal places as needed.) }
\]
Answer
Thus, the probability that a normally distributed random variable $x$ with mean $\mu=89$ and standard deviation $\sigma=5$ is less than 86 is approximately \(\boxed{0.2743}\).
Step 1 ：Given a normally distributed random variable $x$ with mean $\mu=89$ and standard deviation $\sigma=5$, we are asked to find the probability that $x$ is less than 86, denoted as $P(x<86)$.
Step 2 ：We can use the Z-score formula to standardize $x$. The Z-score is given by $Z = \frac{x - \mu}{\sigma}$, where $x$ is the value from the dataset, $\mu$ is the mean and $\sigma$ is the standard deviation. The Z-score tells us how many standard deviations an element is from the mean.
Step 3 ：Substituting the given values into the Z-score formula, we get $Z = \frac{86 - 89}{5} = -0.6$.
Step 4 ：We can then use a Z-table or a statistical software to find the probability that $x$ is less than 86. This is equivalent to finding the probability that $Z$ is less than -0.6.
Step 5 ：From the Z-table or statistical software, we find that the probability is approximately 0.2743.
Step 6 ：Thus, the probability that a normally distributed random variable $x$ with mean $\mu=89$ and standard deviation $\sigma=5$ is less than 86 is approximately \(\boxed{0.2743}\).
