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Statistics, Data, and Probability I: Drill Set 2, Problem 4. Which of the following statements is true?

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Description:

Statistics, Data, and Probability I: Drill Set 1, Problem 4. How old was the final person to join?

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English Language

Transcript

00:03

Here's a boring shmoop-ily unvoiced question you'll find on the exam...

00:07

At a local high school, the mean age of its ten-member glee club is 16.

00:13

With the start of the new school year, the club is expecting the enrollment of

00:17

five additional new members, with the mean age of the first four new recruits as 14.

00:23

If the larger 15-member club turns out to have a mean age of 15.4,

00:28

how old was the final person to join?

00:36

Here are the possible answers:

00:43

For this problem, we really have to know what mean... means.

00:47

Because it has nothing to do with that seriously unfriendly janitor,

00:50

Spyder, who works the night shift at your high school.

00:54

Mean just means... average.

00:55

Yep, we're dealing with an AVERAGING problem.

00:58

Remember Mrs. Slobovitz' words of wisdom:

01:01

Add up all the numbers, then divide by the number of items you added.

01:04

And get your feet off the desk.

01:06

So this question is asking for an average,

01:08

but there are really 2 parts to the... summing we'll have to do.

01:12

Yeah... we're gonna get sum.

01:15

The first part is the ten 16-year-old glee guys and gals.

01:19

16 times 10 is 160, which is how long they've all on this Earth, combined.

01:24

But that's only the first part. Now come the interlopers.

01:29

The second part tells us that 4 recruits have a mean age of 14.

01:33

Do you care if they were 10, 18, 6, and 22? Nah, not really.

01:38

We just want to solve the problem, we're not looking to start a Little League team.

01:42

We only care that the average age of those 4 recruits is 14.

01:47

So now we have 14 of the 15 already in the bag... age-wise...

01:51

...we know the numbers...

01:52

...so we'll be adding 10 times 16, or 160... to 4 times 14, or 56.

02:03

Just like the grocery story check out lady says, Your subtotal is 216, ma'am.

02:10

Only problem is that we're missing one body.

02:15

Let's just step back a sec for a reality check here.

02:18

What's our average up to this point? Well, we have 14 bodies and a total of 216...

02:23

so if we forgot about the last man standing, we could get the average age thus far if we wanted.

02:28

We have 216 divided by 14 which is 15.42... ish

02:32

That's an interesting number because they tell us that the overall average is 15.4.

02:40

Let's say we totally didn't get how to solve this problem. Like... no clue at all.

02:45

If we've got a calculator, the answer is at our fingertips.

02:48

Isn't it about time we gave up and let the machines win?

02:51

Let's look at the answer choices.

02:53

If the last one was 14, then we'd have:

02:55

216 + 14 = 230

02:58

Then we'd divide by 15 and get 15.33.

03:03

If the last one was 15 we'd have:

03:05

216+15 = 231

03:08

Then we'd divide by 15 and get 15.4. Hm...we'll come back to you, buddy...

03:13

If the last one was 16 we'd have:

03:15

216+16= 232

03:17

Then we'd divide by 15 and get 15.466.

03:22

If the last one was 17 we'd have:

03:23

216+17 = 233

03:26

Then we'd divide by 15 and get 15.533.

03:30

The formal name for this method of solving this kind of problem?

03:34

BRUTE FORCE.

03:35

And yeah - it's a tough way to make a living. This result leaves the age of the last recruit,

03:39

which we assume to be x.

03:41

The total sum of all 15 members' ages is 15.4 times 15 or 77 fifths times 15, which equals

03:51

231... so the equation in terms of the data sums is x + 160 + 56 = 231.

03:58

160 plus 56 is 216. If we subtract 216 from both sides, we're left with x on the left

04:04

side and 15 on the right side... so the age of the last member is 15.

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