Answer to Question #197247 in Statistics and Probability for Ezekiel

Question

A module is over assume that the total Mark's were at an average of u=18 per assessment and that the distribution of total Mark's are normally distributed with o=10.

a)what is the proportion that a student would have a mark more than 24 mark if randomly selected?

b)what proportion of students would have marks between 10 and 24?

Show all calculations!

Answer 1

$\mu = 18 \\ \sigma = 10$

a)

$P(X>24) = 1-P(X<24) \\ = 1 -P(Z< \frac{24-18}{10}) \\ = 1 -P(Z< 0.6) \\ = 1 -0.7257 \\ = 0.2743$

b)

$P(10<X<24) = P(X<24) -P(X<10) \\ = P(Z< \frac{24-18}{10}) -P(Z< \frac{10-18}{10}) \\ = P(Z<0.6) -P(Z< -0.8) \\ = 0.7257 -0.2118 \\ = 0.5139$