Answer to Question #267037 in Statistics and Probability for Mayyy

Question

The time a student takes to arrive at a solution for a statistics problem depends upon whether he or she recognizes certain simplifying comments in the problem statement. The probability of this recognition is 0.7. If the student recognizes the comments, the solution time is normally distributed with a mean time of 20 minutes and standard deviation of 4.3 minutes. If the student does not recognize the simplifying comments, the solution time is normally distributed with a mean time of 43 minutes with a standard deviation of 10.2 minutes. a) What is the expected solution time in a large class of students? b) What is the probability that a student chosen at random will require more than 28.2 minutes? c) What is the probability that he or she will require more than 43 minutes?

Answer 1

a) The expected solution time here is computed as:

= E(X | recognition) P( recognition ) + E(X | no recognition ) P( no recognition )

= 20*0.7 + 43*0.3

= 26.9 minutes

b) The probability here is computed as:

P(X > 28.2 )

= P(X > 28.2 | recognition ) P( recognition )+ P( X > 28.2 | no recognition ) P( no recognition )

= P(Z > (28.2 - 20) / 4.3) *0.7 + P(Z > (28.2 - 43) / 10.2) *0.3

= P( Z > 1.91)*0.7 + P(Z > -1.45)*0.3

From standard normal tables,

= 0.0281*0.7 + 0.9265*0.3

= 0.2976

Therefore 0.2976 is the required probability here.

c) The probability here is computed as:

= P(Z > (43 - 20) / 4.3) *0.7 + P(Z > (43 - 43) / 10.2) *0.3

= P(Z > 5.34)*0.7 + P(Z > 0)*0.3

= 0 + 0.5*0.3

= 0.15

Therefore 0.15 is the required probability here.