Probability that a normally distributed random variable with a mean of 8.5 and a standard deviation of 2 is greater than 10

Given: Mean(μ) = 8.5,  Standard Deviation(σ) = 2,  Data Point(X ) = 10
If X  is a normally distributed random variable X~N(μ, σ²), then   is N(0, 1)
Since Z , the standard normal distribution is also N(0, 1), we standardize X  by finding the Z-score corresponding to X  = 10:
Z = = = = 0.75
Therefore, the Probability that X  is greater than 10 equals the Probability that Z  is greater than 0.75:
P (X > 10) = P (Z > 0.75) = 1 – P (Z < 0.75)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.75) = 0.77337
As a result, the Probability that X, a normally distributed random variable with a mean of 8.5 and a standard deviation of 2 is greater than 10:
P (X > 10) = P (Z > 0.75) = 1 – 0.77337 = 0.22663
P (X > 10) = 0.22663 = 22.663%