Question

Two college roommates have each committed to donating to charity each week for the next year. The roommates’ weekly incomes are independent of each other. Suppose the amount donated in a week by one roommate is approximately normal with mean $30 and standard deviation $10, and the amount donated in a week by the other roommate is approximately normal with mean $60 and standard deviation $20. Which of the following is closest to the expected number of weeks in a 52-week year that their combined donation will exceed $120 ?0; the combined donation never exceeds $120 in a weekA1 weekB3 weeksC5 weeksD8 weeks

Answer 1

Answer:

C. 5 weeks.

Step-by-step explanation:

In this question we have a random variable that is equal to the sum of two normal-distributed random variables.

If we have two random variables X and Y, both normally distributed, the sum will have this properties:

$S=X+Y\\\\\ \mu_S=\mu_X+\mu_Y=30+60=90\\\\\sigma_S=\sqrt{\sigma_X^2+\sigma_Y^2}=\sqrt{10^2+20^2}=\sqrt{100+400}=\sqrt{500}=22.36$

To calculate the expected weeks that the donation exceeds $120, first we can calculate the probability of S>120:

$z=\frac{S-\mu_S}{\sigma_S} =\frac{120-90}{22.36}=\frac{30}{22.36}=1.34\\\\P(S>120)=P(z>1.34)=0.09012$

The expected weeks can be calculated as the product of the number of weeks in the year (52) and this probability:

$E=\#weeks*P(S>120)=52*0.09012=4.68$

The nearest answer is C. 5 weeks.

Answer 2

The expected number of weeks can be calculated by finding the probability. To find the probability, we have random variable that is equal to the sum of two normal-distributed random variables.

The correct answer is 5 weeks (c).

Given:

Normal distribution is $\mu_x =30$ and $\mu_y=60$.

Standard deviation is $\sigma_x=10$ and $\sigma_y=20$.

Calculate the normal distribution.

$\mu_s=\mu_x+\mu_y\\\mu_s=30+60\\\mu_s=90$

Calculate the standard deviation.

$\sigma_s=\sqrt{\sigma_{x}^{2}+\sigma_{y}^{2}}\\\sigma_s=\sqrt{10^2+20^2}\\\sigma_s=\sqrt{100+400}\\\sigma_s=22.36$

Calculate the probability of $S>120$.

$z=\dfrac{S-\mu_s}{\sigma_s}\\z=\dfrac{120-90}{22.36}\\z=1.34$

So,

$P(S>120)=P(z>1.34)=0.09012$

Calculate the expected week.

$E=\rm Weeks \times P(S>2)\\E=52\times 0.09012\\E=4.68\\E\approx5$

Thus, the correct answer is 5 weeks (c).

Learn more about what normal-distributed random variables is here:

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