Given: Mean(μ) = 500, Standard Deviation(σ) = 50, Data Point(X1) = 600, Data Point(X2) = 650
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 = 600 and X2 = 650:
Z600 =
Z650 =
Therefore, the Probability that X is between 600 and 650 equals the Probability that Z is between 2.00 and 3.00:
P (600 < X < 650) = P (2.00 < Z < 3.00) = P (Z < 3.00) – P (Z < 2.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 3.00) = 0.99865
P (Z < 2.00) = 0.97725
As a result, the Probability that X, a normally distributed random variable with a mean of 500 and a standard deviation of 50 is between 600 and 650:
P (600 < X < 650) = P (2.00 < Z < 3.00) = 0.99865 – 0.97725 = 0.0214
P (600 < X < 650) = 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 = 600 and X2 = 650:
Z600 =
X1 – μ/σ
= 600 – 500/50
= 100/50
= 2.00Z650 =
X2 – μ/σ
= 650 – 500/50
= 150/50
= 3.00Therefore, the Probability that X is between 600 and 650 equals the Probability that Z is between 2.00 and 3.00:
P (600 < X < 650) = P (2.00 < Z < 3.00) = P (Z < 3.00) – P (Z < 2.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 3.00) = 0.99865
P (Z < 2.00) = 0.97725
As a result, the Probability that X, a normally distributed random variable with a mean of 500 and a standard deviation of 50 is between 600 and 650:
P (600 < X < 650) = P (2.00 < Z < 3.00) = 0.99865 – 0.97725 = 0.0214
P (600 < X < 650) = 0.0214 = 2.14%
