Statistics (math)

[Vastor]

hey guys,

 

the question says...

 

A set of numbers x, 7, 11, 5x has a mean of 9. Find the values of x and the first quartile.

 

firstly, I got x = 3, which is right. but I got problems with finding the first quartile. btw, the answer is 5.

 

so, what does need to be done to find the first quartile in an ungroup data, or does the answer given is wrong??? find quartile in group data seems to be easier... ;O

[rktpro]

hey guys,

 

the question says...

 

A set of numbers x, 7, 11, 5x has a mean of 9. Find the values of x and the first quartile.

 

firstly, I got x = 3, which is right. but I got problems with finding the first quartile. btw, the answer is 5.

 

so, what does need to be done to find the first quartile in an ungroup data, or does the answer given is wrong??? find quartile in group data seems to be easier... ;O

 

Hint:Sometimes a quartile is the average of two numbers which are in the quartile.

[Vastor]

Hint:Sometimes a quartile is the average of two numbers which are in the quartile.

 

so, (3+7) / 2 = 5

 

does that all needed to find every first / third quartile in un-group data?

[rktpro]

so, (3+7) / 2 = 5

 

does that all needed to find every first / third quartile in un-group data?

 

Data must be in ascending order.

[Vastor]

hey guys,

 

The table below shows a set of numbers arranged in descending order where n is a positive integer.

 

 

(a) Express the median of the set of numbers in terms of n. (DONE)

 

(b) Find the possible value of n. (HOW?)

 

© Using the values of n obtained in (b), find the possible value of the mode (this may be solved when (b) solved)

[imatfaal]

Because the question states that they are in descending order you have a set of inequalities for n i.e. n+2>6 etc

[Vastor]

hey guys,

 

The table below shows the distribution of marks obtained by a group of students in a Physics test.

 

 

Construct a cumulative frequency table and draw its ogive, Using the ogive, estimate

(a) the median mark,

(b) the upper quartile mark,

© the percentage of students who obtained more than 60 marks

 

I do calculation rather than construct the cumulative frequency table and draw the ogive, but the answer I got is 49.875, the answer given is (51)

same happen for (b) question, I got 70 instead of 65.5("real" answer)

 

does calculation missed that much from the ogive itself?

[rktpro]

hey guys,

 

The table below shows the distribution of marks obtained by a group of students in a Physics test.

 

 

Construct a cumulative frequency table and draw its ogive, Using the ogive, estimate

(a) the median mark,

(b) the upper quartile mark,

© the percentage of students who obtained more than 60 marks

 

I do calculation rather than construct the cumulative frequency table and draw the ogive, but the answer I got is 49.875, the answer given is (51)

same happen for (b) question, I got 70 instead of 65.5("real" answer)

 

does calculation missed that much from the ogive itself?

 

It wouldn't take enough time if you draw the ogive too. You will yourself understand your question.

[Vastor]

Hey guys,

 

 

The table above shows the distribution of families with the number of children in the familes. Find

(a) the range (done = 4)

(b) the interquartile range (done = 3.5)

© the variance (nope)

(d) the standard deviation (nope)

 

http://www.mathsisfu...-deviation.html

 

conceptually it's understood! but, why: for Variance of a set of ungrouped data

 

[math] \sigma^2 = \frac{\sum(x-\overline x)^2}{N} [/math]

 

or

 

[math]\frac{\sum x^2}{N} - (\frac{\sum x}{N})^2[/math]

 

what is the difference? in my book it tells nothing. -..-"

 

when I tried the second formula it's wrong, for the first formula, I don't know how to use, what is summation symbol precedence compare to multiply/divide?

Edited March 12, 2012 by Vastor

[AJS]

There is no appreciable difference between the two formulas you mention above.

---The first in the definitional formula

--- the second is a computational formula (not frequently used thanks to advances in computational speed)

 

The first formula can be worded verbally as:

The variance is equal to the sum of the squared deviations around the mean divided by sample size.

 

To use it

1. subtract the mean from each individual data point (this will provide you the deviation of that score from the mean)

2. Square each deviation

3. add them all together

4. divide by you total number of data points (if your class has talked about degrees of freedom, you should use N-1 as the denominator instead of just N)

[rktpro]

Vastor, for standard deviation, as I understand, you have to take any larger mean and subtract all other from it.

 

EDIT: For example, say you take a larger mean x, then the standard deviation for the first frequency would be x-(mean of the frequency)

Edited March 12, 2012 by rktpro

[AJS]

For the standard deviation. Just square root of the variance (you computed with the formula)

[AJS]

For the standard deviation. Just square root of the variance (you computed with the formula)

[Vastor]

There is no appreciable difference between the two formulas you mention above.

---The first in the definitional formula

--- the second is a computational formula (not frequently used thanks to advances in computational speed)

 

The first formula can be worded verbally as:

The variance is equal to the sum of the squared deviations around the mean divided by sample size.

 

To use it

1. subtract the mean from each individual data point (this will provide you the deviation of that score from the mean)

2. Square each deviation

3. add them all together

4. divide by you total number of data points (if your class has talked about degrees of freedom, you should use N-1 as the denominator instead of just N)

 

k, lemme try

 

[math]\sigma^2 = \frac{\sum(x-\overline x)^2}{N}[/math]

 

[math]\sigma^2 = \frac{\sum(x-2)^2}{21}[/math]

 

from here, where I assume "x = individucal data point", can you clarify more what is this?

 

I mean the "Number of children per family" or "number of families" etc...

[Vastor]

EDIT: Aww, nvm

 

edit2:

 

AJS, should have put

 

3. multiply with frequency

4. add them all together... etc...

 

I'm confused out of your word actually....

Edited March 13, 2012 by Vastor

[Vastor]

Hey guys,

 

how do you calculate composite index if there are no weightage in the first place?

 

well, let's see the question

 

12. The table below shows the monthly expenditure of Ramli's family in the year 1992 and the price index for the year 1995 using 1992 as the base year

 

Calculate the composite index number for the monthly expenditure of Ramli's family, correct to the nearest integer.

 

I'm trying like this:

 

[math] CI = \frac{130 + 125 + 110 + 105}{4} [math] for assuming that each one has 1 weightage

 

unfortunately, it doesn't work.... the real answer = 122, while my answer = 117.5

 

thnx for anyone help!

 

P.S. actually I thought it should be "the given answer in error", but then it's not when I'm wrong again for question 13. =s

[imatfaal]

I really don't know this part of maths - but this is what I interpret what you are saying.

 

In 1995 you would need to pay 130 for a basket of Food which would have cost you 100 in 1992: this is the food index, we also have housing, transport and other indexes - these are community wide. But we are asked for an individual index for the Ramli family. We cannot just simply average the 4 indexes (which is what you have done) because the Ramlis spend far more on food than they do on transport etc - thus we weight the community indexes (as given in table) by the amounts of each sector that that Ramli family buy.

 

I presume you can do a weighted average using the family expenditure as the weights (BTW the answer does come out as 122) - if you cannot let us know.

[Vastor]

I really don't know this part of maths - but this is what I interpret what you are saying.

 

In 1995 you would need to pay 130 for a basket of Food which would have cost you 100 in 1992: this is the food index, we also have housing, transport and other indexes - these are community wide. But we are asked for an individual index for the Ramli family. We cannot just simply average the 4 indexes (which is what you have done) because the Ramlis spend far more on food than they do on transport etc - thus we weight the community indexes (as given in table) by the amounts of each sector that that Ramli family buy.

 

I presume you can do a weighted average using the family expenditure as the weights (BTW the answer does come out as 122) - if you cannot let us know.

 

hmmm, I got the answer from this:

 

[math] CI = \frac{1995 expenditure}{base(1992) expenditure} * 100 [/math]

 

this show me some new perspective on composite index, thnx for your hint!

 

P.S. "don't know" you say, huh?

[Xin wen]

Hey guys,

 

how do you calculate composite index if there are no weightage in the first place?

 

well, let's see the question

 

12. The table below shows the monthly expenditure of Ramli's family in the year 1992 and the price index for the year 1995 using 1992 as the base year

 

Calculate the composite index number for the monthly expenditure of Ramli's family, correct to the nearest integer.

 

I'm trying like this:

 

[math] CI = \frac{130 + 125 + 110 + 105}{4} [math] for assuming that each one has 1 weightage

 

unfortunately, it doesn't work.... the real answer = 122, while my answer = 117.5

 

thnx for anyone help!

 

P.S. actually I thought it should be "the given answer in error", but then it's not when I'm wrong again for question 13. =s