-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
$$8 \le h < 10\ \ \ \ \ \ \ \ \ \ 6$$
$$10 \le h < 12\ \ \ \ \ \ \ \ 5$$
$$12 \le h < 14\ \ \ \ \ \ \ \ 7 $$
$$14 \le h < 16\ \ \ \ \ \ \ \ 4$$

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

 

[karush]

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

 

[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
$$8 \le h < 10\ \ \ \ \ \ \ \ \ \ 6$$
$$10 \le h < 12\ \ \ \ \ \ \ \ 5$$
$$12 \le h < 14\ \ \ \ \ \ \ \ 7 $$
$$14 \le h < 16\ \ \ \ \ \ \ \ 4$$

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

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
$$\frac{4+5+6+7}{4} = \frac{11}{2} $$

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

 

[karush]

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

so would the median of this be from

$$\{9,11,13,15\} = 12$$

and the mode be $$7$$ 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

$$\{9,11,13,15\} = 12$$

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 $$7$$ 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 $$12\leq h < 14$$ 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, $$\sigma$$) shows how much variation or dispersion exists from the average (mean), or expected value.

So I would presume $$\sigma$$ 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 $$12\leq h < 14$$ 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, $$\sigma$$) shows how much variation or dispersion exists from the average (mean), or expected value.

So I would presume $$\sigma$$ 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 $$\sum n_i $$ would this mean $$6+5+7+4=22$$

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

 

[I like Serena]

by $$\sum n_i $$ would this mean $$6+5+7+4=22$$

Yes!

If so then
$$\frac{6 \times (6-11.82)^2 +5 \times (5-11.82)^2 +7 \times (7-11.82)^2 +4 \times (4-11.82)^2}{22}=variance$$
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?$$\frac{9 \times (9-11.82)^2 +11 \times (11-11.82)^2 +13 \times (13-11.82)^2 +15 \times (15-11.82)^2}{22}=11.307$$ or $$ \sigma^2$$

thus standard deviation would be $$\sqrt{11.307}=3.3626$$

 

[I like Serena]

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

thus standard deviation would be $$\sqrt{11.307}=3.3626$$

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.

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

so if correct then $$\sqrt{4.60331} = 2.14553$$ or $$\sigma$$

 

[I like Serena]

Yep. That looks right.

 

[karush]

there's still more ??

Number of Data
$$= 22$$ assume sum of frequenciesInterquartile range
assume we could go off the intervals

so $$Q_1=10 \ \ Q_2=12 \ \ Q_3=14$$

then $$14-10=4$$

range
$$16-8=8$$

 

[I like Serena]

there's still more ??

Number of Data
$$= 22$$ assume sum of frequencies

Correct.

Interquartile range
assume we could go off the intervals

so $$Q_1=10 \ \ Q_2=12 \ \ Q_3=14$$

then $$14-10=4$$

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
$$16-8=8$$

Right.

 

[karush]

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