Given: Mean(μ) = 650, Standard Deviation(σ) = 25, Data Point(X1) = 600, Data Point(X2) = 660
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 = 660:
Z600 =
Z660 =
Therefore, the Probability that X is between 600 and 660 equals the Probability that Z is between -2.00 and 0.40:
P (600 < X < 660) = P (-2.00 < Z < 0.40) = P (Z < 0.40) – P (Z < -2.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.40) = 0.65542
P (Z < -2.00) = P (Z > 2.00) = 1 – P (Z < 2.00) = 1 – 0.97725 = 0.02275
As a result, the Probability that X, a normally distributed random variable with a mean of 650 and a standard deviation of 25 is between 600 and 660:
P (600 < X < 660) = P (-2.00 < Z < 0.40) = 0.65542 – 0.02275 = 0.63267
P (600 < X < 660) = 0.63267 = 63.267%
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 = 660:
Z600 =
X1 – μ/σ
= 600 – 650/25
= -50/25
= -2.00Z660 =
X2 – μ/σ
= 660 – 650/25
= 10/25
= 0.40Therefore, the Probability that X is between 600 and 660 equals the Probability that Z is between -2.00 and 0.40:
P (600 < X < 660) = P (-2.00 < Z < 0.40) = P (Z < 0.40) – P (Z < -2.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.40) = 0.65542
P (Z < -2.00) = P (Z > 2.00) = 1 – P (Z < 2.00) = 1 – 0.97725 = 0.02275
As a result, the Probability that X, a normally distributed random variable with a mean of 650 and a standard deviation of 25 is between 600 and 660:
P (600 < X < 660) = P (-2.00 < Z < 0.40) = 0.65542 – 0.02275 = 0.63267
P (600 < X < 660) = 0.63267 = 63.267%
