Given: Mean(μ) = 6, Standard Deviation(σ) = 2, Data Point(X1) = 5.54, Data Point(X2) = 6.53
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 = 5.54 and X2 = 6.53:
Z5.54 =
Z6.53 =
Therefore, the Probability that X is between 5.54 and 6.53 equals the Probability that Z is between -0.23 and 0.27:
P (5.54 < X < 6.53) = P (-0.23 < Z < 0.27) = P (Z < 0.27) – P (Z < -0.23)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.27) = 0.60642
P (Z < -0.23) = P (Z > 0.23) = 1 – P (Z < 0.23) = 1 – 0.59095 = 0.40905
As a result, the Probability that X, a normally distributed random variable with a mean of 6 and a standard deviation of 2 is between 5.54 and 6.53:
P (5.54 < X < 6.53) = P (-0.23 < Z < 0.27) = 0.60642 – 0.40905 = 0.19737
P (5.54 < X < 6.53) = 0.19737 = 19.737%
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 = 5.54 and X2 = 6.53:
Z5.54 =
X1 – μ/σ
= 5.54 – 6/2
= -0.46/2
= -0.23Z6.53 =
X2 – μ/σ
= 6.53 – 6/2
= 0.53/2
= 0.27 (rounded to 2 decimal places)Therefore, the Probability that X is between 5.54 and 6.53 equals the Probability that Z is between -0.23 and 0.27:
P (5.54 < X < 6.53) = P (-0.23 < Z < 0.27) = P (Z < 0.27) – P (Z < -0.23)
From a standard normal distribution table (values rounded to five decimal places):
P (Z < 0.27) = 0.60642
P (Z < -0.23) = P (Z > 0.23) = 1 – P (Z < 0.23) = 1 – 0.59095 = 0.40905
As a result, the Probability that X, a normally distributed random variable with a mean of 6 and a standard deviation of 2 is between 5.54 and 6.53:
P (5.54 < X < 6.53) = P (-0.23 < Z < 0.27) = 0.60642 – 0.40905 = 0.19737
P (5.54 < X < 6.53) = 0.19737 = 19.737%
