-aug.16.data list from table intervals

[karush]

in creating a list of numbers to find mean, median, mode, range and some other questions with this table

height in meters| frequency
\(\displaystyle 8 \le h < 10\ \ \ \ \ \ \ \ \ \ 6\)
\(\displaystyle 10 \le h < 12\ \ \ \ \ \ \ \ 5\)
\(\displaystyle 12 \le h < 14\ \ \ \ \ \ \ \ 7 \)
\(\displaystyle 14 \le h < 16\ \ \ \ \ \ \ \ 4\)

how do we make a list when you have intervals? or do you just use the number in between like \(\displaystyle 8 \le h < 10\) would be \(\displaystyle \{9,9,9,9,9,9\}\)

 

[karush]

let me ask a different question would the mean of this table be
\(\displaystyle \frac{4+5+6+7}{4} = \frac{11}{2} \)

 

[I like Serena]

Hi karush! :)

in creating a list of numbers to find mean, median, mode, range and some other questions with this table

height in meters| frequency
\(\displaystyle 8 \le h < 10\ \ \ \ \ \ \ \ \ \ 6\)
\(\displaystyle 10 \le h < 12\ \ \ \ \ \ \ \ 5\)
\(\displaystyle 12 \le h < 14\ \ \ \ \ \ \ \ 7 \)
\(\displaystyle 14 \le h < 16\ \ \ \ \ \ \ \ 4\)

how do we make a list when you have intervals? or do you just use the number in between like \(\displaystyle 8 \le h < 10\) would be \(\displaystyle \{9,9,9,9,9,9\}\)

Yes. You would use the number in the middle as you suggest.

let me ask a different question would the mean of this table be
\(\displaystyle \frac{4+5+6+7}{4} = \frac{11}{2} \)

So no.
The mean would be \(\displaystyle \frac{6\cdot 9 + 5 \cdot 11 + 7 \cdot 13 + 4 \cdot 15}{6+5+7+4}\).

 

[karush]

The mean would be \(\displaystyle \frac{6\cdot 9 + 5 \cdot 11 + 7 \cdot 13 + 4 \cdot 15}{6+5+7+4}\).

so would the median of this be from

\(\displaystyle \{9,11,13,15\} = 12\)

and the mode be \(\displaystyle 7\) since it has the highest frequency

do I have to start a new OP if I continue to ask more Q on this table?

 

[MarkFL]

...
do I have to start a new OP if I continue to ask more Q on this table?

As long as your additional questions pertain to the data already provided, it is best to ask further questions regarding it here in this topic. :D

 

[I like Serena]

so would the median of this be from

\(\displaystyle \{9,11,13,15\} = 12\)

The median is the height where half is smaller and the other half is taller.
At height 12, you have 6+5=11 people smaller, and 7+4=11 people taller.
So indeed the median is 12.

and the mode be \(\displaystyle 7\) since it has the highest frequency

The mode is the height that occurs most... but no one has height 7. ;)

 

[karush]

The mode is the height that occurs most... but no one has height 7.

so the most frequent is \(\displaystyle 12\leq h < 14\) or 13 for mode.

my next question is standard deviation and variance
from Wikipedia
In statistics and probability theory, standard deviation (represented by the symbol sigma, \(\displaystyle \sigma\)) shows how much variation or dispersion exists from the average (mean), or expected value.

So I would presume \(\displaystyle \sigma\) here is 2 since that is the size of the intervals

I read variance in Wikipedia but not sure if it applies to this table.
so how is variance derived?

 

[I like Serena]

so the most frequent is \(\displaystyle 12\leq h < 14\) or 13 for mode.

Right. :)

my next question is standard deviation and variance
from Wikipedia
In statistics and probability theory, standard deviation (represented by the symbol sigma, \(\displaystyle \sigma\)) shows how much variation or dispersion exists from the average (mean), or expected value.

So I would presume \(\displaystyle \sigma\) here is 2 since that is the size of the intervals

I read variance in Wikipedia but not sure if it applies to this table.
so how is variance derived?

Not quite.

Let's start with variance.
In your case the formula is:
$$\text{Variance} = \frac{\sum n_i \times (x_i - \text{mean})^2}{\sum n_i}$$
where $n_i$ is the frequency of each category, $x_i$ is the mid value of each height interval, and $\text{mean}$ is the value you already found.

Variance is often denoted as $\sigma^2$.
Standard deviation (denoted as $\sigma$) is the square root of the variance.

 

[karush]

Let's start with variance.
In your case the formula is:
$$\text{Variance} = \frac{\sum n_i \times (x_i - \text{mean})^2}{\sum n_i}$$
where $n_i$ is the frequency of each category, $x_i$ is the mid value of each height interval, and $\text{mean}$ is the value you already found.

Variance is often denoted as $\sigma^2$.
Standard deviation (denoted as $\sigma$) is the square root of the variance.

by \(\displaystyle \sum n_i \) would this mean \(\displaystyle 6+5+7+4=22\)

If so then
\(\displaystyle \frac{6 \times (6-11.82)^2 +5 \times (5-11.82)^2 +7 \times (7-11.82)^2 +4 \times (4-11.82)^2}{22}=variance\)
or is this composed wrong?

 

[I like Serena]

by \(\displaystyle \sum n_i \) would this mean \(\displaystyle 6+5+7+4=22\)

Yes!

If so then
\(\displaystyle \frac{6 \times (6-11.82)^2 +5 \times (5-11.82)^2 +7 \times (7-11.82)^2 +4 \times (4-11.82)^2}{22}=variance\)
or is this composed wrong?

Almost.
But you have substituted the frequencies instead of the mid interval values for $x_i$.

 

[karush]

Almost.
But you have substituted the frequencies instead of the mid interval values for $x_i$.

how this?\(\displaystyle \frac{9 \times (9-11.82)^2 +11 \times (11-11.82)^2 +13 \times (13-11.82)^2 +15 \times (15-11.82)^2}{22}=11.307\) or \(\displaystyle \sigma^2\)

thus standard deviation would be \(\displaystyle \sqrt{11.307}=3.3626\)

 

[I like Serena]

how this?\(\displaystyle \frac{9 \times (9-11.82)^2 +11 \times (11-11.82)^2 +13 \times (13-11.82)^2 +15 \times (15-11.82)^2}{22}=11.307\) or \(\displaystyle \sigma^2\)

thus standard deviation would be \(\displaystyle \sqrt{11.307}=3.3626\)

Hold on.
Now you have substituted the mid interval values for the freqencies $n_i$.
Check where it says $n_i$ and where it says $x_i$.

Btw, the meaning of variance is the average of the squared deviations from the mean.

 

[karush]

Hold on.
Now you have substituted the mid interval values for the freqencies $n_i$.
Check where it says $n_i$ and where it says $x_i$.

Btw, the meaning of variance is the average of the squared deviations from the mean.

\(\displaystyle \frac{6 \times (9-11.82)^2 +5 \times (11-11.82)^2 +7 \times (13-11.82)^2 +4 \times (15-11.82)^2}{22}=4.60331 \) or \(\displaystyle \sigma^2\)

so if correct then \(\displaystyle \sqrt{4.60331} = 2.14553\) or \(\displaystyle \sigma\)

 

[I like Serena]

Yep. That looks right.

 

[karush]

there's still more ??

Number of Data
\(\displaystyle = 22\) assume sum of frequenciesInterquartile range
assume we could go off the intervals

so \(\displaystyle Q_1=10 \ \ Q_2=12 \ \ Q_3=14\)

then \(\displaystyle 14-10=4\)

range
\(\displaystyle 16-8=8\)

 

[I like Serena]

there's still more ??

Number of Data
\(\displaystyle = 22\) assume sum of frequencies

Correct.

Interquartile range
assume we could go off the intervals

so \(\displaystyle Q_1=10 \ \ Q_2=12 \ \ Q_3=14\)

then \(\displaystyle 14-10=4\)

You're not supposed to work from the intervals.
$Q_1$ is the height such that 25 percent is below.
Since 25% of 22 persons is 5.5, the $Q_1$ height is somewhere in the interval 8-10, which contains 6 persons.
There can be some discussion where that height actually is when talking about intervals, but let's keep it simple and say that $Q_1=9$, which is the middle of the lowest interval.
Similarly $Q_3$ is the height with 75% below.
Keeping it simple that is the middle of the third interval. So $Q_1=13$.

Anyway, your interquartile range comes out the same.

range
\(\displaystyle 16-8=8\)

Right.

 

[karush]

thanks everyone for your help. it was a new topic for me
sure I'll be back with more