Question

A professor of statistics noticed that the marks in his course are normally distributed. He also noticed that his morning classes average 71% with a standard deviation of 12% on their final exams. His afternoon classes average 78% with a standard deviation of 8%. A. What is the probability that a randomly selected student in the morning class has a higher final exam mark than a randomly selected student from an afternoon class? Probability = B. What is the probability that the mean mark of four randomly selected students from a morning class is greater than the average mark of four randomly selected students from an afternoon class? Probability =

Answer 1

Answer:

0.3137 ; 0.2228

Step-by-step explanation:

Given a normal distribution :

Morning class :

Mean(Mm) = 71%

Standard deviation (Sm) = 12%

Afternoon class:

Mean(Ma) = 78%

Standard deviation (Sa) = 8%

M = Mm - Ma = (71 - 78) = - m7

S = √Sm + Sa = √12² + 8² = √208

A. What is the probability that a randomly selected student in the morning class has a higher final exam mark than a randomly selected student from an afternoon class?

P(morning > afternoon) = p(morning - afternoon > 0)

Using:

Z = (0 - (-7)) / S

Z = 7 / √208

Z = 0.4853628

P(Z > 0.49) = 0.3137

B)

What is the probability that the mean mark of four randomly selected students from a morning class is greater than the average mark of four randomly selected students from an afternoon class?

Using:

Z = (4 - (-7)) / S

Z = 11 / √208

Z = 0.7627127

P(Z > 0.49) = 0.2228