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Mathematics 12 Online
OpenStudy (vwrgwegw):

find the mean

OpenStudy (vwrgwegw):

OpenStudy (sweetemy):

It's 3 since you round it up to the nearest tenth, and the answer is 2.5.

OpenStudy (vwrgwegw):

so am i right?

OpenStudy (sweetemy):

Yes I would think so, there is .5 meaning you have to round up.

OpenStudy (vwrgwegw):

ok can u help me on one more?

OpenStudy (sweetemy):

I can try!

OpenStudy (vwrgwegw):

Leah predicts that an average of 5 passengers will board the bus at each of the 6 stops along her route. The numbers of passengers that board at the first five stops are 3, 7, 5, 6, and 2. How many passengers need to board at the sixth stop for her prediction to be true? A.3.8 B.4.6 C.5 D.7

OpenStudy (vwrgwegw):

do i ave to add ALL the numbers and then divide or just the passangers and the stops and divide for 5? or like how?

OpenStudy (sweetemy):

Okay so you have to add 3, 7, 5, 6, 2 AND one of the numbers from the answer bank, then divide that by 6. Do that till you get 5

OpenStudy (vwrgwegw):

ok doing tht will get me the answer then right? :D

OpenStudy (sweetemy):

Yes, whichever one gets you the mean of 5 is the right answer,

OpenStudy (vwrgwegw):

7

OpenStudy (sweetemy):

Yes, that's what I got too.

OpenStudy (vwrgwegw):

THANK U SO MUCH!!!!

OpenStudy (sweetemy):

No problem!

OpenStudy (mathstudent55):

Look at the first question again. Your answer is incorrect.

OpenStudy (sweetemy):

Answer is 3 without rounding.

OpenStudy (mathstudent55):

No. How many numbers did you average? 5 numbers? The numbers are 1, 2, 3, 4, 5, and you averaged them and got 3?

OpenStudy (sweetemy):

Oh no, it's for the value, I thought we were doing it for the frequency. That's my mistake then, sorry.

OpenStudy (mathstudent55):

Notice that they created a problem that if you add the values you get 15, and if you add the frequencies, you also get 15. Then you divide by 5, thinking you are finding the mean. and you get a mean of 3. Both of the above ways are incorrect. The mean is not 3.

OpenStudy (sweetemy):

Okay that's my bad, thank you for pointing that out.

OpenStudy (mathstudent55):

The frequency is important. The table is telling you that the values are not simply one 1, one 2, one 3, one 4, and one 5, and there are 5 different values that you need to average.

OpenStudy (mathstudent55):

The frequency tells you how many of each value you have. There are 7 values of 1. There is 1 value of 2. There are 3 values of 3. There are 2 values of 4. There are 2 values of 5. 7 + 1 + 3 + 2 + 2 = 15 You are averaging 15 values. Let's get the sum of the values: There are 7 values of 1. ---> 7 * 1 = 7 There is 1 value of 2. ---> 1 * 2 = 2 There are 3 values of 3. ---> 3 * 3 = 9 There are 2 values of 4. ---> 2 * 4 = 8 There are 2 values of 5. ---> 2 * 5 = 10 The sum of all 15 values is: 7 + 2 + 9 + 8 + 10 = 36 There is a total of 15 values. The mean is obtained by dividing the sum of the values by the number of values. \(mean = \dfrac{36}{15} = 2.4\)

OpenStudy (mathstudent55):

For the second problem, you are correct. The answer is 7. Your method worked. Here is another way of solving it. You have 6 stops with an average of 5 passengers per stop. That means 5 * 6 = 30 passengers are expected. Now subtract the passengers that already came in from 30, and that will give you what is needed in the 6th stop. 30 - 3 - 7 - 5 - 6 - 2 = 7 You can also add first the passengers from the first 5 stops and subtract that from 30: 30 - (3 + 7 + 5 + 6 + 2) = 30 - 23 = 7

OpenStudy (sweetemy):

I didn't read the question like that. I didn't realize that frequency was included in this, I actually don't remember learning about that. I'm sorry for the error and thank you for showing me this. I thought that it was a straight add and divide question.

OpenStudy (mathstudent55):

You're welcome. This is a reminder for you as to how the frequency is used.

OpenStudy (sweetemy):

Thank you for the help, and the second method is less time consuming.

OpenStudy (mathstudent55):

You're welcome.

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