Given: Mean(μ) = 4, Standard Deviation(σ) = 1, Data Point(X1) = 3, Data Point(X2) = 4
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 = 3 and X2 = 4:
Z3 =
Z4 =
Therefore, the Probability that X is between 3 and 4 equals the Probability that Z is between -1.00 and 0.00:
P (3 < X < 4) = P (-1.00 < Z < 0.00) = P (Z < 0.00) – P (Z < -1.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.00) = 0.5
P (Z < -1.00) = P (Z > 1.00) = 1 – P (Z < 1.00) = 1 – 0.84134 = 0.15866
As a result, the Probability that X, a normally distributed random variable with a mean of 4 and a standard deviation of 1 is between 3 and 4:
P (3 < X < 4) = P (-1.00 < Z < 0.00) = 0.5 – 0.15866 = 0.34134
P (3 < X < 4) = 0.34134 = 34.134%
Note: P (3 < X < 4) is equal to the Probability that X is between the Mean and 1 Standard Deviation below the Mean
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 = 3 and X2 = 4:
Z3 =
X1 – μ/σ
= 3 – 4/1
= -1/1
= -1.00Z4 =
X2 – μ/σ
= 4 – 4/1
= 0/1
= 0.00Therefore, the Probability that X is between 3 and 4 equals the Probability that Z is between -1.00 and 0.00:
P (3 < X < 4) = P (-1.00 < Z < 0.00) = P (Z < 0.00) – P (Z < -1.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.00) = 0.5
P (Z < -1.00) = P (Z > 1.00) = 1 – P (Z < 1.00) = 1 – 0.84134 = 0.15866
As a result, the Probability that X, a normally distributed random variable with a mean of 4 and a standard deviation of 1 is between 3 and 4:
P (3 < X < 4) = P (-1.00 < Z < 0.00) = 0.5 – 0.15866 = 0.34134
P (3 < X < 4) = 0.34134 = 34.134%
Note: P (3 < X < 4) is equal to the Probability that X is between the Mean and 1 Standard Deviation below the Mean
