Thread: statistics problem

Having trouble setting up and solving these 2 word problems.



1. A set of data includes all of the positive odd integers less than 100, the positive, two-digit multiples of 10, and the numbers 4, 16, and 64. All included integers appear exactly once in the data. What is the positive difference between the median and the mean of the set of data? Express your answer as a decimal to the nearest thousandth.


2. Ayushi has six coins with a total value of 30 cents. The coins are not all the same. Two of Ayushi's coins will be chosen at random. What is the probability that the totalvalue of the two coins will be less than 15 cents? express your answer as a common fraction.

Originally Posted by Brent

Having trouble setting up and solving these 2 word problems.



1. A set of data includes all of the positive odd integers less than 100, the positive, two-digit multiples of 10, and the numbers 4, 16, and 64. All included integers appear exactly once in the data. What is the positive difference between the median and the mean of the set of data? Express your answer as a decimal to the nearest thousandth.

Mr F says: Have you tried listing the data and applying the definitions of median and mean to it?


2. Ayushi has six coins with a total value of 30 cents. The coins are not all the same. Two of Ayushi's coins will be chosen at random. What is the probability that the totalvalue of the two coins will be less than 15 cents? express your answer as a common fraction.

Mr F says: Have you tried listing what the six coins might be? Some possibilities are

1, 2, 2, 5, 10, 10

2, 2, 2, 2, 2, 20

etc.

I'd imagine that each combination of coins is equally likely .....

(I've assumed not all the same means that some but not all coins can be the same. If that's not the case then I have no idea what the six coins might be).

..

Originally Posted by mr fantastic

..

1. with #1 i am completely lost in how to set it up to solve. I know its asking for all of the odd intergers less then 100, whick would be.

1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 .... etc!

Then all multipuls of 10 which would be... 10,20,30 etc...

And last 4,16,64

All of these intergers only appear once.

As far as the median and the mean i'm totally lost with this, it is something i have yet to be taught.


2. I'm not sure if this problem is a ratio problem, i know the 6 coins have to be (5) pennys and (1) quarter. But 2 coins are being drawn each time so i'm not sure if the probally of drawing less then 15 cents would be expressed as 5/6 or since 2 coins are drawn it would be expressed as 3/6

Originally Posted by Brent

1. with #1 i am completely lost in how to set it up to solve. I know its asking for all of the odd intergers less then 100, whick would be.

1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 .... etc!

Then all multipuls of 10 which would be... 10,20,30 etc...

And last 4,16,64

All of these intergers only appear once.

As far as the median and the mean i'm totally lost with this, it is something i have yet to be taught.


2. I'm not sure if this problem is a ratio problem, i know the 6 coins have to be (5) pennys and (1) quarter. But 2 coins are being drawn each time so i'm not sure if the probally of drawing less then 15 cents would be expressed as 5/6 or since 2 coins are drawn it would be expressed as 3/6

Mean: Add all the numbers up and divide by how many there are.
Median: Write the numbers out from smallest to largest. The middle one is the median.

2. Calculate the probability of choosing two pennies. A tree diagram might help. I get (5/6)(4/5) = 20/30.