Probability that a normally distributed random variable with a mean of 5 and a standard deviation of 1 is less than 4 or greater than 6

Given: Mean(μ) = 5,  Standard Deviation(σ) = 1,  Data Point(X₁) = 4,  Data Point(X₂) = 6
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-scores corresponding to X₁ = 4 and X₂ = 6:
Z₄ = = = = -1.00
Z₆ = = = = 1.00
Therefore, the Probability that X  is less than 4 or greater than 6 equals the Probability that Z  is less than -1.00 or greater than 1.00:
P (X < 4) + P (X > 6) = P (Z < -1.00) + P (Z > 1.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < -1.00) = P (Z > 1.00) = 1 – P (Z < 1.00) = 1 – 0.84134 = 0.15866
P (Z > 1.00) = 1 – P (Z < 1.00) = 1 – 0.84134 = 0.15866
As a result, the Probability that X, a normally distributed random variable with a mean of 5 and a standard deviation of 1 is less than 4 or greater than 6:
P (X < 4) + P (X > 6) = P (Z < -1.00) + P (Z > 1.00) = 0.15866 + 0.15866 = 0.31732
P (X < 4) + P (X > 6) = 0.31732 = 31.732%
Note: P (X < 4) + P (X > 6) is equal to the Probability that X  is more than 1 Standard Deviation away from the Mean