Assume the random variable $\mathrm{x}$ is normally distributed with mean $\mu=86$ and standard deviation $\sigma=4$. Find the indicated probability.
\[
P(76< x< 81)
\]
\[
P(76< x< 81)=
\]
(Round to four decimal places as needed.)

Step 1 ：Given a normally distributed random variable x with mean \(\mu = 86\) and standard deviation \(\sigma = 4\), we are asked to find the probability that x falls between 76 and 81, i.e., \(P(76 < x < 81)\).

Step 2 ：We first convert the given x-values into z-scores using the formula \(z = \frac{x - \mu}{\sigma}\).

Step 3 ：For x = 76, the z-score \(z_1\) is calculated as \(z_1 = \frac{76 - 86}{4} = -2.5\).

Step 4 ：For x = 81, the z-score \(z_2\) is calculated as \(z_2 = \frac{81 - 86}{4} = -1.25\).

Step 5 ：We then find the probabilities corresponding to these z-scores using the standard normal distribution. The probability corresponding to \(z_1 = -2.5\) is approximately 0.0062, and the probability corresponding to \(z_2 = -1.25\) is approximately 0.1056.

Step 6 ：The probability that the random variable x falls between 76 and 81 is the difference between these two probabilities, i.e., \(P = p_2 - p_1 = 0.1056 - 0.0062 = 0.0994\).

Step 7 ：Final Answer: The probability that the random variable x falls between 76 and 81 is approximately \(\boxed{0.0994}\).