Given: Mean(μ) = 5, Standard Deviation(σ) = 4, Data Point(X ) = 6
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 = 6:
Z =
Therefore, the Probability that X is greater than 6 equals the Probability that Z is greater than 0.25:
P (X > 6) = P (Z > 0.25) = 1 – P (Z < 0.25)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.25) = 0.59871
As a result, the Probability that X, a normally distributed random variable with a mean of 5 and a standard deviation of 4 is greater than 6:
P (X > 6) = P (Z > 0.25) = 1 – 0.59871 = 0.40129
P (X > 6) = 0.40129 = 40.129%
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 = 6:
Z =
X – μ/σ
= 6 – 5/4
= 1/4
= 0.25Therefore, the Probability that X is greater than 6 equals the Probability that Z is greater than 0.25:
P (X > 6) = P (Z > 0.25) = 1 – P (Z < 0.25)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.25) = 0.59871
As a result, the Probability that X, a normally distributed random variable with a mean of 5 and a standard deviation of 4 is greater than 6:
P (X > 6) = P (Z > 0.25) = 1 – 0.59871 = 0.40129
P (X > 6) = 0.40129 = 40.129%
