Given: Mean(μ) = 5, Standard Deviation(σ) = 1, Data Point(X ) = 7
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-score corresponding to X = 7:
Z =
Therefore, the Probability that X is less than 7 equals the Probability that Z is less than 2.00:
P (X < 7) = P (Z < 2.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 2.00) = 0.97725
As a result, the Probability that X, a normally distributed random variable with a mean of 5 and a standard deviation of 1 is less than 7:
P (X < 7) = P (Z < 2.00) = 0.97725
P (X < 7) = 0.97725 = 97.725%
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-score corresponding to X = 7:
Z =
X – μ/σ
= 7 – 5/1
= 2/1
= 2.00Therefore, the Probability that X is less than 7 equals the Probability that Z is less than 2.00:
P (X < 7) = P (Z < 2.00)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 2.00) = 0.97725
As a result, the Probability that X, a normally distributed random variable with a mean of 5 and a standard deviation of 1 is less than 7:
P (X < 7) = P (Z < 2.00) = 0.97725
P (X < 7) = 0.97725 = 97.725%
