Given: Mean(μ) = 2, Standard Deviation(σ) = 2, 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 less than 2 equals the Probability that Z is less than 0.00:
P (X < 2) = P (Z < 0.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.00) = 0.5
As a result, the Probability that X, a normally distributed random variable with a mean of 2 and a standard deviation of 2 is less than 2:
P (X < 2) = P (Z < 0.00) = 0.5
P (X < 2) = 0.5 = 50%
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 – 2/2
= 0/2
= 0.00Therefore, the Probability that X is less than 2 equals the Probability that Z is less than 0.00:
P (X < 2) = P (Z < 0.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.00) = 0.5
As a result, the Probability that X, a normally distributed random variable with a mean of 2 and a standard deviation of 2 is less than 2:
P (X < 2) = P (Z < 0.00) = 0.5
P (X < 2) = 0.5 = 50%
