Given: Mean(μ) = 35, Standard Deviation(σ) = 6, Data Point(X ) = 42
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 = 42:
Z =
Therefore, the Probability that X is less than 42 equals the Probability that Z is less than 1.17:
P (X < 42) = P (Z < 1.17)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 1.17) = 0.879
As a result, the Probability that X, a normally distributed random variable with a mean of 35 and a standard deviation of 6 is less than 42:
P (X < 42) = P (Z < 1.17) = 0.879
P (X < 42) = 0.879 = 87.9%
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 = 42:
Z =
X – μ/σ
= 42 – 35/6
= 7/6
= 1.17 (rounded to 2 decimal places)Therefore, the Probability that X is less than 42 equals the Probability that Z is less than 1.17:
P (X < 42) = P (Z < 1.17)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 1.17) = 0.879
As a result, the Probability that X, a normally distributed random variable with a mean of 35 and a standard deviation of 6 is less than 42:
P (X < 42) = P (Z < 1.17) = 0.879
P (X < 42) = 0.879 = 87.9%
