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%