Let x be a continuous random variable that follows a normal distribution with a mean of 240 and a standard deviation of 51. Find the value of x so that the area under the normal curve to the right of x is approximately 0.3512. Round your answer to two decimal places

To find the value of x such that the area under the normal curve to the right of x is approximately 0.3512, we need to use the z-score formula:

z = \dfrac{x - \mu}{\sigma}

where:
x = the value we are looking for,
\mu = 240 (mean),
\sigma = 51 (standard deviation).

First, we need to find the z-score corresponding to the given area using the standard normal distribution table:

P(Z > z) = 0.3512

Since the standard normal table gives the area to the left of z, we need to find the z-score corresponding to the area to the left of our desired area:

P(Z < z) = 1 - 0.3512 = 0.6488

Looking up 0.6488 in the z-table gives us a z-score of about 0.37.

Now, we can plug the z-score into the z-score formula and solve for x:

0.37 = \frac{x - 240}{51}

0.37 * 51 = x - 240

x = 0.37 * 51 + 240

x = 18.87 + 240

x \approx 258.87

Therefore, the value of x such that the area under the normal curve to the right of x is approximately 0.3512 is approximately 258.87.

\boxed{x \approx 258.87}