Suppose a normal distribution has a mean of 98 and a standard deviation of 6. What is P(x ≥ 86)?

To find P(x ≥ 86), we first need to convert the raw score (x) of 86 to a z-score. The z-score is a measure of how many standard deviations an element is from the mean.

The formula for calculating the z-score is:

Z = (X - μ) / σ

where: X = raw score μ = mean σ = standard deviation

Substituting the given values into the formula, we get:

Z = (86 - 98) / 6 = -2

So, the z-score for X = 86 is -2.

Next, we need to find the probability that the z-score is greater than or equal to -2. This is the same as finding the area to the right of z = -2 under the standard normal curve.

Looking up z = -2 in the standard normal distribution table or using a calculator, we find that the area to the left of z = -2 is approximately 0.0228.

However, we want the area to the right of z = -2, so we subtract the area to the left from 1 (since the total area under the curve is 1):

P(x ≥ 86) = 1 - 0.0228 = 0.9772

So, the probability that x is greater than or equal to 86 is approximately 0.9772, or 97.72%.