Given: Mean(μ) = 0, Standard Deviation(σ) = 1, Data Point(X1) = -1.96, Data Point(X2) = 1.96
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 = -1.96 and X2 = 1.96:
Z-1.96 =
Z1.96 =
Therefore, the Probability that X is between -1.96 and 1.96 equals the Probability that Z is between -1.96 and 1.96:
P (-1.96 < X < 1.96) = P (-1.96 < Z < 1.96) = P (Z < 1.96) – P (Z < -1.96)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 1.96) = 0.975
P (Z < -1.96) = P (Z > 1.96) = 1 – P (Z < 1.96) = 1 – 0.975 = 0.025
As a result, the Probability that X, a normally distributed random variable with a mean of 0 and a standard deviation of 1 is between -1.96 and 1.96:
P (-1.96 < X < 1.96) = P (-1.96 < Z < 1.96) = 0.975 – 0.025 = 0.95
P (-1.96 < X < 1.96) = 0.95 = 95%
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 = -1.96 and X2 = 1.96:
Z-1.96 =
X1 – μ/σ
= -1.96 – 0/1
= -1.96/1
= -1.96Z1.96 =
X2 – μ/σ
= 1.96 – 0/1
= 1.96/1
= 1.96Therefore, the Probability that X is between -1.96 and 1.96 equals the Probability that Z is between -1.96 and 1.96:
P (-1.96 < X < 1.96) = P (-1.96 < Z < 1.96) = P (Z < 1.96) – P (Z < -1.96)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 1.96) = 0.975
P (Z < -1.96) = P (Z > 1.96) = 1 – P (Z < 1.96) = 1 – 0.975 = 0.025
As a result, the Probability that X, a normally distributed random variable with a mean of 0 and a standard deviation of 1 is between -1.96 and 1.96:
P (-1.96 < X < 1.96) = P (-1.96 < Z < 1.96) = 0.975 – 0.025 = 0.95
P (-1.96 < X < 1.96) = 0.95 = 95%
