Given: Mean(μ) = 6, Standard Deviation(σ) = 5, Data Point(X1) = 6, Data Point(X2) = 12
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 = 6 and X2 = 12:
Z6 =
Z12 =
Therefore, the Probability that X is between 6 and 12 equals the Probability that Z is between 0.00 and 1.20:
P (6 < X < 12) = P (0.00 < Z < 1.20) = P (Z < 1.20) – P (Z < 0.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 1.20) = 0.88493
P (Z < 0.00) = 0.5
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 6 and 12:
P (6 < X < 12) = P (0.00 < Z < 1.20) = 0.88493 – 0.5 = 0.38493
P (6 < X < 12) = 0.38493 = 38.493%
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 = 6 and X2 = 12:
Z6 =
X1 – μ/σ
= 6 – 6/5
= 0/5
= 0.00Z12 =
X2 – μ/σ
= 12 – 6/5
= 6/5
= 1.20Therefore, the Probability that X is between 6 and 12 equals the Probability that Z is between 0.00 and 1.20:
P (6 < X < 12) = P (0.00 < Z < 1.20) = P (Z < 1.20) – P (Z < 0.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 1.20) = 0.88493
P (Z < 0.00) = 0.5
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 6 and 12:
P (6 < X < 12) = P (0.00 < Z < 1.20) = 0.88493 – 0.5 = 0.38493
P (6 < X < 12) = 0.38493 = 38.493%
