Professor Blockhus gives two different statistics tests, but one test is harder than the other. Scores on test A are normally distributed with a mean score of 78 and a standard deviation of 6. Scores on test B are also normally distributed but with a mean score of 65 and a standard deviation of 9. If Karl scored an 85 on test A, what percent of the class scored below him

87.83%

Step-by-step explanation:

Solution:-

- Denote a random variable "X" denoting the scores on test A. The random variable follows a normal distribution with parameters mean ( μ ) and standard deviation ( σ ) as follows:

                           X ~ Norm ( μ , σ^2 )

                           X ~ Norm ( 78 , 6^2 ).

- Karl takes the test A and scores 85 on the test. The percent of people who scored below him can be defined by:

                           P ( X < 85 )

- We will standardize our test value and compute the Z-score:

                           P ( Z < ( x - μ )  / σ )

Where, x : The test value

                           P (  Z < ( 85 - 78 )  / 6 )

                           P (  Z < 1.1667 )

- Then use the Z-standardize tables for the following probability:

                           P ( Z < 1.1667 ) = 0.8783

Therefore,          P ( X < 85 ) = 0.8783

- The percentage of student who scored below karl were:

                          = 100*P ( X < 85 ) = 100*0.8783

                          = 87.83%

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: