Question

normally distributed with mean of 18 and standard deviation of 6 determine the probability of students with 33 or higher

Answer 1

Given:

Mean, = 18

Standard deviation, = 6

X ~ Normal (=18, ^2 = 6^2)

For normal distribution :

z-score = X - /

The probability of students with 33 or higher :

P(X 33) = 1 - P(X < 33)

= 1 - P((X-)/ < (33-18)/6)

= 1 - P( Z < 2.5)

= 1 - 0.9938

= 0.0062

P(X 33) = 0.0062

The probability of students with 33 or higher is 0.0062