Thread: Mean and Median

Hello

This looks a really easy question but i dont actually know the answer. Could someone please tell me how i work it out

The five number summary for the weights (in pounds) of fish caught in a bass tournament is:

Min: 2.3
Q1: 2.8
Medium: 3.0
Q3: 3.3
Max: 4.5

Would you expect the mean weight of all the fish caught to be higher or lower than the median? Explain

Thanks

Originally Posted by Stem01

Hello

This looks a really easy question but i dont actually know the answer. Could someone please tell me how i work it out

The five number summary for the weights (in pounds) of fish caught in a bass tournament is:

Min: 2.3
Q1: 2.8
Medium: 3.0
Q3: 3.3
Max: 4.5

Would you expect the mean weight of all the fish caught to be higher or lower than the median? Explain

Thanks

Do you understand what "mean" and "median" are? The mean is what you might know as "arithmetic average": add all the weights together, then divide by the number of fish. The median is just the middle number, disregarding how far apart the numbers might be. I notice that the distance from the minimum weight to the middle weight is 3.0-2.3= .7 pounds while the distance from the middle weight to the maximum weight is 4.5-3.0= 1.5 pounds, twice the previous distance. So the weights tend to be "stretched" toward the high side. Remember what I said about the median disregarding things like that?

2.3
2.8
3.0
3.3
4.5

=15.9

So 15.9/5 = 3.18 which is above the mean, so it means that it is above the median?

I have no idea what that calculation is suppose to represent. You say it is "above the mean". Do you know what the mean is? How did you find it?

Originally Posted by Stem01

2.3
2.8
3.0
3.3
4.5

=15.9

So 15.9/5 = 3.18 which is above the mean, so it means that it is above the median?

The mean is NOT calculated by adding up the numbers from the five number summary and dividing by 5!! It's impossible to calculate the mean from the given information. The best that can be done is what the question is asking you to do. Please review post #2.

Originally Posted by Stem01

Hello

This looks a really easy question but i dont actually know the answer. Could someone please tell me how i work it out

The five number summary for the weights (in pounds) of fish caught in a bass tournament is:

Min: 2.3
Q1: 2.8
Medium: 3.0
Q3: 3.3
Max: 4.5

Would you expect the mean weight of all the fish caught to be higher or lower than the median? Explain

Thanks

MEDIAN:The Median may be defined as that value which divides a distribution into two parts such that an exactly equal number of scores fall above and below the point.
In other words 50 percent of the scores will be above the median and the remaining 50 percent below it.
finding the mean value for the above data given as follows:
Here I am calculating the mean by Calculating the mean from frequency distribution.
Class interval :1-3 ;Mid Value(x):2 Frequency(f):4(2.3,2.8,3.0,3.3);fx=8
Class interval:4-6 ; Mid value(x):5 :
Frequency(f):1(4.5);fx=5
fx = 8+5=13
Frequency(f)=4+1=5
Mean value=summation fx /summation f = 13/5 = 2.6
and the median for this data 2.3,2.8,3,3.3,4.5=3
Therefore, here the mean is less than the median.

Originally Posted by Stem01

Hello

This looks a really easy question but i dont actually know the answer. Could someone please tell me how i work it out

The five number summary for the weights (in pounds) of fish caught in a bass tournament is:

Min: 2.3
Q1: 2.8
Medium: 3.0
Q3: 3.3
Max: 4.5

Would you expect the mean weight of all the fish caught to be higher or lower than the median? Explain

Thanks

What are Q1 and Q3?
At first I thought they were quartiles.

Originally Posted by saranya

MEDIAN:The Median may be defined as that value which divides a distribution into two parts such that an exactly equal number of scores fall above and below the point.
In other words 50 percent of the scores will be above the median and the remaining 50 percent below it.
finding the mean value for the above data given as follows:
Here I am calculating the mean by Calculating the mean from frequency distribution.
Class interval :1-3 ;Mid Value(x):2 Frequency(f):4(2.3,2.8,3.0,3.3);fx=8
Class interval:4-6 ; Mid value(x):5 :
Frequency(f):1(4.5);fx=5
fx = 8+5=13
Frequency(f)=4+1=5
Mean value=summation fx /summation f = 13/5 = 2.6
and the median for this data 2.3,2.8,3,3.3,4.5=3
Therefore, here the mean is less than the median.

Sorry, but what you've posted makes no sense. It's clearly stated that the data is a five number summary. Q1 etc. are quartiles. They are not the weights of the fish in the sample.