Question

Let X be normally distributed with mean μ = 2,800 and standard deviation σ = 1,300. Find x such that P(X ≤ x) = 0.9382. (Round "z" value to 2 decimal places, and final answer to nearest whole number.)

Answer 1

To find the value of x such that P(X ≤ x) = 0.9382, you can find the z-score corresponding to this probability and then use the formula z = (x - μ) / σ to find x. Plugging in the values, x = 5524.

Step-by-step explanation:

To find the value of x such that P(X ≤ x) = 0.9382, we need to find the z-score corresponding to this probability and then use the formula z = (x - μ) / σ to find x.

First, let's find the z-score using a standard normal distribution table or calculator. A z-score of 0.9382 corresponds to a value of approximately 1.88.

Now, we can rearrange the z-score formula to solve for x: x = z * σ + μ. Plugging in the values, we get:

x = 1.88 * 1300 + 2800 = 5524.

Therefore, x can be rounded to the nearest whole number, which is 5524.