Assume the random variable $x$ is normally distributed with mean $\mu=50$ and standard deviation $\sigma=7$. Find the indicated probability. \[ P(x>34) \] \[ P(x>34)= \] (Round to four decimal places as needed.)

Step 1 ：We are given a normally distributed random variable $x$ with mean $\mu=50$ and standard deviation $\sigma=7$. We are asked to find the probability that $x$ is greater than 34, i.e., $P(x>34)$.

Step 2 ：To solve this, we first standardize the value 34 to a z-score. The z-score represents how many standard deviations away from the mean the value is. The formula for calculating a z-score is $z = \frac{x - \mu}{\sigma}$, where $x$ is the value we're interested in (34 in this case), $\mu$ is the mean (50), and $\sigma$ is the standard deviation (7).

Step 3 ：Substituting the given values into the formula, we get $z = \frac{34 - 50}{7} = -2.2857142857142856$.

Step 4 ：After calculating the z-score, we can use a z-table to find the probability that a random variable from a standard normal distribution (mean = 0, standard deviation = 1) is greater than the calculated z-score. This probability is the same as the probability that the original random variable is greater than 34.

Step 5 ：From the z-table, we find that the probability that the random variable $x$ is greater than 34 is approximately 0.9889. This means that there is a 98.89% chance that a randomly selected value of $x$ will be greater than 34. This makes sense given that 34 is less than the mean of 50, so we would expect the majority of values to be greater than 34.

Step 6 ：Final Answer: The probability that the random variable $x$ is greater than 34 is approximately \(\boxed{0.9889}\).