Given: Mean(μ) = 5, Standard Deviation(σ) = 1, Data Point(X1) = 2, 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 = 2 and X2 = 3:
Z2 =
Z3 =
Therefore, the Probability that X is between 2 and 3 equals the Probability that Z is between -3.00 and -2.00:
P (2 < X < 3) = P (-3.00 < Z < -2.00) = P (Z < -2.00) – P (Z < -3.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < -2.00) = P (Z > 2.00) = 1 – P (Z < 2.00) = 1 – 0.97725 = 0.02275
P (Z < -3.00) = P (Z > 3.00) = 1 – P (Z < 3.00) = 1 – 0.99865 = 0.00135
As a result, the Probability that X, a normally distributed random variable with a mean of 5 and a standard deviation of 1 is between 2 and 3:
P (2 < X < 3) = P (-3.00 < Z < -2.00) = 0.02275 – 0.00135 = 0.0214
P (2 < X < 3) = 0.0214 = 2.14%
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 = 2 and X2 = 3:
Z2 =
X1 – μ/σ
= 2 – 5/1
= -3/1
= -3.00Z3 =
X2 – μ/σ
= 3 – 5/1
= -2/1
= -2.00Therefore, the Probability that X is between 2 and 3 equals the Probability that Z is between -3.00 and -2.00:
P (2 < X < 3) = P (-3.00 < Z < -2.00) = P (Z < -2.00) – P (Z < -3.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < -2.00) = P (Z > 2.00) = 1 – P (Z < 2.00) = 1 – 0.97725 = 0.02275
P (Z < -3.00) = P (Z > 3.00) = 1 – P (Z < 3.00) = 1 – 0.99865 = 0.00135
As a result, the Probability that X, a normally distributed random variable with a mean of 5 and a standard deviation of 1 is between 2 and 3:
P (2 < X < 3) = P (-3.00 < Z < -2.00) = 0.02275 – 0.00135 = 0.0214
P (2 < X < 3) = 0.0214 = 2.14%
