Given: Mean(μ) = 6, Standard Deviation(σ) = 5, Data Point(X1) = 0, Data Point(X2) = 8
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 = 8:
Z0 =
Z8 =
Therefore, the Probability that X is between 0 and 8 equals the Probability that Z is between -1.20 and 0.40:
P (0 < X < 8) = P (-1.20 < Z < 0.40) = P (Z < 0.40) – P (Z < -1.20)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.40) = 0.65542
P (Z < -1.20) = P (Z > 1.20) = 1 – P (Z < 1.20) = 1 – 0.88493 = 0.11507
As a result, the Probability that X, a normally distributed random variable with a mean of 6 and a standard deviation of 5 is between 0 and 8:
P (0 < X < 8) = P (-1.20 < Z < 0.40) = 0.65542 – 0.11507 = 0.54035
P (0 < X < 8) = 0.54035 = 54.035%
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 = 8:
Z0 =
X1 – μ/σ
= 0 – 6/5
= -6/5
= -1.20Z8 =
X2 – μ/σ
= 8 – 6/5
= 2/5
= 0.40Therefore, the Probability that X is between 0 and 8 equals the Probability that Z is between -1.20 and 0.40:
P (0 < X < 8) = P (-1.20 < Z < 0.40) = P (Z < 0.40) – P (Z < -1.20)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.40) = 0.65542
P (Z < -1.20) = P (Z > 1.20) = 1 – P (Z < 1.20) = 1 – 0.88493 = 0.11507
As a result, the Probability that X, a normally distributed random variable with a mean of 6 and a standard deviation of 5 is between 0 and 8:
P (0 < X < 8) = P (-1.20 < Z < 0.40) = 0.65542 – 0.11507 = 0.54035
P (0 < X < 8) = 0.54035 = 54.035%
