Given: Mean(μ) = 0, Standard Deviation(σ) = 1, Data Point(X ) = 2
If X is a normally distributed random variable X~N(μ, σ2), then
Since Z , the standard normal distribution is also N(0, 1), we standardize X by finding the Z-score corresponding to X = 2:
Z =
Therefore, the Probability that X is greater than 2 equals the Probability that Z is greater than 2.00:
P (X > 2) = P (Z > 2.00) = 1 – P (Z < 2.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 2.00) = 0.97725
As a result, the Probability that X, a normally distributed random variable with a mean of 0 and a standard deviation of 1 is greater than 2:
P (X > 2) = P (Z > 2.00) = 1 – 0.97725 = 0.02275
P (X > 2) = 0.02275 = 2.275%
Note: P (X > 2) is equal to the Probability that X is more than 2 Standard Deviations above the Mean
If X is a normally distributed random variable X~N(μ, σ2), then
X – μ/σ
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 = 2:
Z =
X – μ/σ
= 2 – 0/1
= 2/1
= 2.00Therefore, the Probability that X is greater than 2 equals the Probability that Z is greater than 2.00:
P (X > 2) = P (Z > 2.00) = 1 – P (Z < 2.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 2.00) = 0.97725
As a result, the Probability that X, a normally distributed random variable with a mean of 0 and a standard deviation of 1 is greater than 2:
P (X > 2) = P (Z > 2.00) = 1 – 0.97725 = 0.02275
P (X > 2) = 0.02275 = 2.275%
Note: P (X > 2) is equal to the Probability that X is more than 2 Standard Deviations above the Mean
