Given: Mean(μ) = 6, Standard Deviation(σ) = 5, Data Point(X ) = 21
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 = 21:
Z =
Therefore, the Probability that X is greater than 21 equals the Probability that Z is greater than 3.00:
P (X > 21) = P (Z > 3.00) = 1 – P (Z < 3.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 3.00) = 0.99865
As a result, the Probability that X, a normally distributed random variable with a mean of 6 and a standard deviation of 5 is greater than 21:
P (X > 21) = P (Z > 3.00) = 1 – 0.99865 = 0.00135
P (X > 21) = 0.00135 = 0.135%
Note: P (X > 21) is equal to the Probability that X is more than 3 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 = 21:
Z =
X – μ/σ
= 21 – 6/5
= 15/5
= 3.00Therefore, the Probability that X is greater than 21 equals the Probability that Z is greater than 3.00:
P (X > 21) = P (Z > 3.00) = 1 – P (Z < 3.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 3.00) = 0.99865
As a result, the Probability that X, a normally distributed random variable with a mean of 6 and a standard deviation of 5 is greater than 21:
P (X > 21) = P (Z > 3.00) = 1 – 0.99865 = 0.00135
P (X > 21) = 0.00135 = 0.135%
Note: P (X > 21) is equal to the Probability that X is more than 3 Standard Deviations above the Mean
