Question

In 2016, the mean score on the AP Art History exam was 3.8, with a standard deviation of 1.2. The mean score on the AP English exam was a 2.9, with a standard deviation of 1.7. What is the combined mean and standard deviation for these exams?
A.) Mean, 0.9; standard deviation, 0.2
B.) Mean, 3.35; standard deviation, 2.9
C.) Mean, 3.35; standard deviation, 1.45
D.) Mean, 6.7; standard deviation, 2.08
E.) Mean, 6.7; standard deviation, 4.33

Answer 1

Answer:

$X+Y \sim N(\mu_X +\mu_Y , \sqrt{\sigma^2_X +\sigma^2_Y})$

The mean is given by:

$\mu = 3.8+2.9= 6.7$

And the standard deviation would be:

$\sigma =\sqrt{1.2^2 + 1.7^2} = 2.08$

And the distribution for X+Y would be:

$X+Y \sim N(6.7 , 2.08)$

And the best answer would be:

D.) Mean, 6.7; standard deviation, 2.08

Step-by-step explanation:

Let X the random variable who represent the AP Art History exam we know that the distribution for X is given by:

$X \sim N(3.8, 1.2)$

Let Y the random variable who represent the AP English exam we know that the distribution for X is given by:

$X \sim N(2.9, 1.7)$

We want to find the distribution for X+Y. Assuming independence between the two distributions we have:

$X+Y \sim N(\mu_X +\mu_Y , \sqrt{\sigma^2_X +\sigma^2_Y})$

The mean is given by:

$\mu = 3.8+2.9= 6.7$

And the standard deviation would be:

$\sigma =\sqrt{1.2^2 + 1.7^2} = 2.08$

And the distribution for X+Y would be:

$X+Y \sim N(6.7 , 2.08)$

And the best answer would be:

D.) Mean, 6.7; standard deviation, 2.08

Answer 2

Answer: D

Step-by-step explanation: