Given: Mean(μ) = 2, Standard Deviation(σ) = 4, Data Point(X1) = -1, Data Point(X2) = 5
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-scores corresponding to X1 = -1 and X2 = 5:
Z-1 =
Z5 =
Therefore, the Probability that X is less than -1 or greater than 5 equals the Probability that Z is less than -0.75 or greater than 0.75:
P (X < -1) + P (X > 5) = P (Z < -0.75) + P (Z > 0.75)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < -0.75) = P (Z > 0.75) = 1 – P (Z < 0.75) = 1 – 0.77337 = 0.22663
P (Z > 0.75) = 1 – P (Z < 0.75) = 1 – 0.77337 = 0.22663
As a result, the Probability that X, a normally distributed random variable with a mean of 2 and a standard deviation of 4 is less than -1 or greater than 5:
P (X < -1) + P (X > 5) = P (Z < -0.75) + P (Z > 0.75) = 0.22663 + 0.22663 = 0.45326
P (X < -1) + P (X > 5) = 0.45326 = 45.326%
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-scores corresponding to X1 = -1 and X2 = 5:
Z-1 =
X1 – μ/σ
= -1 – 2/4
= -3/4
= -0.75Z5 =
X2 – μ/σ
= 5 – 2/4
= 3/4
= 0.75Therefore, the Probability that X is less than -1 or greater than 5 equals the Probability that Z is less than -0.75 or greater than 0.75:
P (X < -1) + P (X > 5) = P (Z < -0.75) + P (Z > 0.75)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < -0.75) = P (Z > 0.75) = 1 – P (Z < 0.75) = 1 – 0.77337 = 0.22663
P (Z > 0.75) = 1 – P (Z < 0.75) = 1 – 0.77337 = 0.22663
As a result, the Probability that X, a normally distributed random variable with a mean of 2 and a standard deviation of 4 is less than -1 or greater than 5:
P (X < -1) + P (X > 5) = P (Z < -0.75) + P (Z > 0.75) = 0.22663 + 0.22663 = 0.45326
P (X < -1) + P (X > 5) = 0.45326 = 45.326%
