Final answer:
To determine the probability of x exceeding 115 in a normal distribution with μ = 100 and σ = 10, we calculate the z-score and then use a standard normal distribution table to find that the probability is approximately 0.0668 or 6.68%.
Step-by-step explanation:
To find the probability that a normally distributed variable x exceeds 115 when μ = 100 and σ = 10, we first convert the value 115 to a z-score. The z-score formula is:
Z = (x - μ) / σ
Plugging the given values in:
Z = (115 - 100) / 10 = 1.5
Now, we need to find the probability that Z is greater than 1.5 using a standard normal distribution table or a calculator. Since we are looking for the area to the right (tail) of the z-score, we need to consider the upper-tail probability.
Looking up the z-score of 1.5 on the standard normal distribution table, we find that the area to the left under the curve is approximately 0.9332. However, since we need the area to the right (greater than), we subtract this value from 1:
P(X > 115) = 1 - P(Z ≤ 1.5)
P(X > 115) = 1 - 0.9332
P(X > 115) = 0.0668
So the probability that x exceeds 115 is roughly 0.0668 or 6.68%, rounded to the nearest hundredth.