Given: Mean(μ) = 7.5, Standard Deviation(σ) = 1.25, Data Point(X ) = 6.5
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 = 6.5:
Z =
Therefore, the Probability that X is less than 6.5 equals the Probability that Z is less than -0.80:
P (X < 6.5) = P (Z < -0.80) = P (Z > 0.80) = 1- P (Z < 0.80)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.80) = 0.78814
As a result, the Probability that X, a normally distributed random variable with a mean of 7.5 and a standard deviation of 1.25 is less than 6.5:
P (X < 6.5) = P (Z < -0.80) = P (Z > 0.80) = 1- 0.78814 = 0.21186
P (X < 6.5) = 0.21186 = 21.186%
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 = 6.5:
Z =
X – μ/σ
= 6.5 – 7.5/1.25
= -1/1.25
= -0.80Therefore, the Probability that X is less than 6.5 equals the Probability that Z is less than -0.80:
P (X < 6.5) = P (Z < -0.80) = P (Z > 0.80) = 1- P (Z < 0.80)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.80) = 0.78814
As a result, the Probability that X, a normally distributed random variable with a mean of 7.5 and a standard deviation of 1.25 is less than 6.5:
P (X < 6.5) = P (Z < -0.80) = P (Z > 0.80) = 1- 0.78814 = 0.21186
P (X < 6.5) = 0.21186 = 21.186%
