Given: Mean(μ) = 2, Standard Deviation(σ) = 4, Data Point(X1) = 1, Data Point(X2) = 3
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 = 3:
Z1 =
Z3 =
Therefore, the Probability that X is between 1 and 3 equals the Probability that Z is between -0.25 and 0.25:
P (1 < X < 3) = P (-0.25 < Z < 0.25) = P (Z < 0.25) – P (Z < -0.25)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.25) = 0.59871
P (Z < -0.25) = P (Z > 0.25) = 1 – P (Z < 0.25) = 1 – 0.59871 = 0.40129
As a result, the Probability that X, a normally distributed random variable with a mean of 2 and a standard deviation of 4 is between 1 and 3:
P (1 < X < 3) = P (-0.25 < Z < 0.25) = 0.59871 – 0.40129 = 0.19742
P (1 < X < 3) = 0.19742 = 19.742%
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 = 3:
Z1 =
X1 – μ/σ
= 1 – 2/4
= -1/4
= -0.25Z3 =
X2 – μ/σ
= 3 – 2/4
= 1/4
= 0.25Therefore, the Probability that X is between 1 and 3 equals the Probability that Z is between -0.25 and 0.25:
P (1 < X < 3) = P (-0.25 < Z < 0.25) = P (Z < 0.25) – P (Z < -0.25)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.25) = 0.59871
P (Z < -0.25) = P (Z > 0.25) = 1 – P (Z < 0.25) = 1 – 0.59871 = 0.40129
As a result, the Probability that X, a normally distributed random variable with a mean of 2 and a standard deviation of 4 is between 1 and 3:
P (1 < X < 3) = P (-0.25 < Z < 0.25) = 0.59871 – 0.40129 = 0.19742
P (1 < X < 3) = 0.19742 = 19.742%
