Given: Mean(μ) = 2, Standard Deviation(σ) = 4, Data Point(X1) = 0, Data Point(X2) = 1
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 = 0 and X2 = 1:
Z0 =
Z1 =
Therefore, the Probability that X is between 0 and 1 equals the Probability that Z is between -0.50 and -0.25:
P (0 < X < 1) = P (-0.50 < Z < -0.25) = P (Z < -0.25) – P (Z < -0.50)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < -0.25) = P (Z > 0.25) = 1 – P (Z < 0.25) = 1 – 0.59871 = 0.40129
P (Z < -0.50) = P (Z > 0.50) = 1 – P (Z < 0.50) = 1 – 0.69146 = 0.30854
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 0 and 1:
P (0 < X < 1) = P (-0.50 < Z < -0.25) = 0.40129 – 0.30854 = 0.09275
P (0 < X < 1) = 0.09275 = 9.275%
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 = 0 and X2 = 1:
Z0 =
X1 – μ/σ
= 0 – 2/4
= -2/4
= -0.50Z1 =
X2 – μ/σ
= 1 – 2/4
= -1/4
= -0.25Therefore, the Probability that X is between 0 and 1 equals the Probability that Z is between -0.50 and -0.25:
P (0 < X < 1) = P (-0.50 < Z < -0.25) = P (Z < -0.25) – P (Z < -0.50)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < -0.25) = P (Z > 0.25) = 1 – P (Z < 0.25) = 1 – 0.59871 = 0.40129
P (Z < -0.50) = P (Z > 0.50) = 1 – P (Z < 0.50) = 1 – 0.69146 = 0.30854
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 0 and 1:
P (0 < X < 1) = P (-0.50 < Z < -0.25) = 0.40129 – 0.30854 = 0.09275
P (0 < X < 1) = 0.09275 = 9.275%
