This is a mean with percentage problem...

Mr. Ramirez and Ms. Jonsey gave tests to their classes with the results seen in the table.

| Overall Mean | Mean for Females | Mean for Males | Percent Females |

Ramirez | 218 | 230 | 205 | |

Jonsey | 221 | 224 | | 88 |

What are the missing entries? For an overall mean, Ms. Jonsey’s class is higher, but for females, the mean is higher in Mr. Ramirez’s class. Is it possible that the mean for males is higher in Mr. Ramirez’s class as well?

The hint they give me is:

To find the entry for the empty cell in the bottom row, consider the remaining percentage for Ms. Jonsey's class and how that percentage was used to determine the overall mean. For the missing entry in Mr. Ramirez's class, think algebraically about the conditions on the total percentages and how those percentages were used to obtain the overall mean for his class.

I think the means for Males in Ms. Jonsey's class is 218, but I am not sure. My professor has not answered emails and I am stuck. Can anyone please help me in understanding this problem?

Thank you in advance and I really appreciate the help I receive.

Schuyla

Re: This is a mean with percentage problem...

This looks tricky, the only way I can find a result for that bottom cell is using a weighted mean (don't think that equals the overall mean in this case)

88% female implies 12% male and thus

$\displaystyle \displaystyle 0.88\times 224 + 0.12\times \mu_M = 211 \implies \mu_M = 199.1$

Re: This is a mean with percentage problem...

Re: This is a mean with percentage problem...

I think 211 is a typo for 221

Re: This is a mean with percentage problem...

For the Ramirez group, suppose that there are $\displaystyle f$ females and $\displaystyle m$ males.

The total score for the females will be $\displaystyle 230f$ and for the males$\displaystyle 205m$ leading to an overall mean of

$\displaystyle (230f+205m)/(f+m).$

Put this equal to $\displaystyle 218$ and simplify to arrive at $\displaystyle 12f=13m.$

The percentage number of females in the group will be $\displaystyle 100f/(f+m),$ and on substituting $\displaystyle m=12f/13,$ this works out at $\displaystyle 52\%.$

For the Jonsey group you have to work this calculation in reverse, that is, start with the percentage number of females and then move on to the overall mean.

Re: This is a mean with percentage problem...

Thank you to both of you for your help! I now understand how to solve this problem.

Re: This is a mean with percentage problem...

Let f be the number of females in the class, m the number of males, A the "mean for females" and B the "mean for males".

The total scores for females is Af and the total score for males is Bm. The total of all scores is Af+ Bm. The overall mean, then, is $\displaystyle C= \frac{Af+ Bm}{m+ f}= A\frac{f}{m+f}+ B\frac{f}{m+f}$. Of course, those two fractions are "the percent of females" and "the percent of males": the overall mean is "mean for females times percentage of females"+ "mean for males times percentage of males".

Here's how I would have done the first problem: Let p= "percentage of females" (in the class). Then the percentage of males is 1- p so the overall mean grade is p(230)+ (1- p)(205)= 218. Solve that equation of p.

For the second use the same formula: let m be the mean for males in the class. The females form 88% of the class so males constitute 12% of it. We must have .88(224)+ .12m= 221. Solve that for m.