Given: Mean(μ) = 6, Standard Deviation(σ) = 5, Data Point(X1) = -2, Data Point(X2) = 0
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 = -2 and X2 = 0:
Z-2 =
Z0 =
Therefore, the Probability that X is between -2 and 0 equals the Probability that Z is between -1.60 and -1.20:
P (-2 < X < 0) = P (-1.60 < Z < -1.20) = P (Z < -1.20) – P (Z < -1.60)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < -1.20) = P (Z > 1.20) = 1 – P (Z < 1.20) = 1 – 0.88493 = 0.11507
P (Z < -1.60) = P (Z > 1.60) = 1 – P (Z < 1.60) = 1 – 0.9452 = 0.0548
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 -2 and 0:
P (-2 < X < 0) = P (-1.60 < Z < -1.20) = 0.11507 – 0.0548 = 0.06027
P (-2 < X < 0) = 0.06027 = 6.027%
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 = -2 and X2 = 0:
Z-2 =
X1 – μ/σ
= -2 – 6/5
= -8/5
= -1.60Z0 =
X2 – μ/σ
= 0 – 6/5
= -6/5
= -1.20Therefore, the Probability that X is between -2 and 0 equals the Probability that Z is between -1.60 and -1.20:
P (-2 < X < 0) = P (-1.60 < Z < -1.20) = P (Z < -1.20) – P (Z < -1.60)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < -1.20) = P (Z > 1.20) = 1 – P (Z < 1.20) = 1 – 0.88493 = 0.11507
P (Z < -1.60) = P (Z > 1.60) = 1 – P (Z < 1.60) = 1 – 0.9452 = 0.0548
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 -2 and 0:
P (-2 < X < 0) = P (-1.60 < Z < -1.20) = 0.11507 – 0.0548 = 0.06027
P (-2 < X < 0) = 0.06027 = 6.027%
