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## When calculating to find the mean and standard deviation of these two sets of numbers: (a)4 0 1 4 3 6 (b)5 3 1 3 4 2 Which data set is more spread out?

Shwetank Mauria
Featured 8 months ago

Data-set (a) is more spread out.

#### Explanation:

Both the given data sets have exactly $6$ data points and their sum is too $18$, hence mean of both is $\frac{18}{6} = 3$.

As regards which data is more spread out, there are mainly three measures for the same.

First is Range, which is the difference of larges and smallest data point. As the range of data-set (a) is $6 - 0 = 6$ and that of data-set (b) is $5 - 1 = 4$, hence data-set (a) is more spread out.

Another measure is mean deviation, which is the average of difference (i.e. absolute value) between data point and mean.

In data-set (a) absolute value of deviations (note it makes deviations positive and hence does not cancel out, which will happen if we keep positive or negative sign of differences) is $\left\{| 4 - 3 | , | 0 - 3 | . | 1 - 3 | , | 4 - 3 | , | 3 - 3 | , | 6 - 3 |\right\}$ i.e. $\left\{1 , 3 , 2 , 1 , 0 , 3\right\}$ and their mean is $\frac{1 + 3 + 2 + 1 + 0 + 3}{6} = \frac{10}{6} = \frac{5}{3}$. In data-set (b) absolute value of deviations is $\left\{| 5 - 3 | , | 3 - 3 | . | 1 - 3 | , | 3 - 3 | , | 4 - 3 | , | 2 - 3 |\right\}$ i.e. $\left\{2 , 0 , 2 , 0 , 1 , 1\right\}$ and their mean is $\frac{2 + 0 + 2 + 0 + 1 + 1}{6} = \frac{6}{6} = 1$. As mean deviation in data-set (a) is higher, it is more spread out.

The third measure is standard deviation, for which we square the deviations (again to make them positive), take their average and then take square root.

We have already worked out deviations above. In data-set (a) squares of deviations are$\left\{1 , 9 , 4 , 1 , 0 , 9\right\}$ and their mean is $\frac{1 + 9 + 4 + 1 + 0 + 9}{6} = \frac{24}{6} = 4$ and standard deviation is $\sqrt{4} = 2$. In data-set (b) squares of deviations are $\left\{4 , 0 , 4 , 0 , 1 , 1\right\}$ and their mean is $\frac{4 + 0 + 4 + 0 + 1 + 1}{6} = \frac{10}{6} = \frac{5}{3}$ and standard deviation is $\sqrt{\frac{5}{3}} = 1.291$. Again as standard deviation in data-set (a) is higher, it is more spread out.

## In poker, a 5-card hand is called two pair if there are two cards of one rank, two cards of another rank, and a fifth card of a third order of the cards doesn't matter (so, example, QQ668 and 68QQ6 are the same). How many 5-card hands are two pair?

Parzival S.
Featured 8 months ago

$78 \times 6 \times 6 \times 11 \times 4 = 2 , 598 , 960$

#### Explanation:

Let's work this through in bits.

Ok - first to know is that a standard card deck has 52 cards. There are 13 cards (1 - 10, Jack, Queen, King - let's call them Ordinals) in each of four Suits (spades, clubs, hearts, diamonds), which is $4 \times 13 = 52$

Ok, let's now work through the number of 5 card hands that are 2 pairs.

The first two cards will be the first card each of the 2 pairs. We have 13 ordinals to choose from and we need one, so we can express that as:

#C_(13,2)=(13!)/((2!)(11!))=78#

(I'm using C to mean Combination, the first number being the number of things we can choose from, and the second number being the number being selected). So there are 78 different ways to pick 2 ordinals from a group of 13.

Next thing we need to do is pick 2 cards from each suit for each ordinal. We can write doing it once as:

#C_(4,2)=(4!)/((2!)(2!))=6#

and we need to do it twice, so let's write the formula again for the other suit:

#C_(4,2)=(4!)/((2!)(2!))=6#

So now we have 2 cards in one ordinal and 2 in the other ordinal. Two pairs.

Lastly we need to account for the last card in 5 card hand. There are 11 ordinals to pick from and we need only 1:

#C_(11,1)=(11!)/((10!)(1!))=11#

and from that one ordinal we need 1 suit:

#C_(4,1)=(4!)/((1!)(3!))=4#

To find the number of 2-pair hands, we multiply it all together:

$78 \times 6 \times 6 \times 11 \times 4 = 2 , 598 , 960$

https://en.wikipedia.org/wiki/Poker_probability#Frequency_of_5-card_poker_hands

## What is the total sum of squares?

Geoff K.
Featured 6 months ago

The total sum of squares (also written $S {S}_{\text{Total}}$ or $S {S}_{T}$) is simply the sum of all the squared data values:

$S {S}_{T} = {\sum}_{i} {y}_{i}^{2}$.

#### Explanation:

As written above, if we have $N$ total observations in our data set, then $S {S}_{T}$ has $N$ degrees of freedom (d.f.), because all $N$ variables used in the sum are free to vary.

Sometimes we account for the average of all the data values (that is, $\overline{y}$) by instead squaring the differences between each data point ${y}_{i}$ and the overall average $\overline{y}$. In this case, we write:

$S {S}_{T} = {\sum}_{i} {\left({y}_{i} - \overline{y}\right)}^{2}$

This version of total sum of squares is the corrected total sum of squares, and if we need to clarify that this is different than the uncorrected version, we can denote this something like $S {S}_{\text{Total, Corrected}}$.

$S {S}_{\text{Tot. Cor.}}$ has one fewer degree of freedom than $S {S}_{T}$. This is because all but one of the $N$ differences (that is, ${y}_{i} - \overline{y} ,$ for $i = 1 , \ldots , N$) are free to vary.

e.g. Suppose I tell you I have three numbers whose average is 4. How many of those numbers do I have to tell you in order for you to know all of them? The answer is "one less than all 3"; because you know the average, you can use the first two numbers I give you to figure out the third number by using

$\overline{y} = \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3} \iff {y}_{3} = 3 \overline{y} - \left({y}_{1} + {y}_{2}\right)$.

If the first two numbers are 3 and 4, you know the last number is 5. If I give you 2 and 4, you know the last number is 6. In this sense, one of the three data points is not free to vary. And since we are using the (fixed) average $\overline{y}$ in calculating $S {S}_{\text{Tot. Cor.}} ,$ it only has $N - 1$ degrees of freedom.

## What is the probability of sitting 5 girls and 5 boys alternatively in a row ?

dk_ch
Featured 2 months ago

The arragement of sitting of 5 Boys and 5 Girls alternatively in a row may start with either a Boy or a Girl. So 2 types of starting are possible.

$\text{Type I} \to B G B G B G B G B G$

$\text{Typy II} \to G B G B G B G B G B$

In each type 5 Boys and 5 Girls may take their positions in #5!# ways. So total number of possible arrangements becomes

#=2xx5!xx5!#

and this is the number of favourable events.

Again without restriction Boys and Girls may be arranged randomly in #10!# ways which is possible number of events.

So the probability of sitting in alternative manner will be given by
$P = \text{favourable number of events"/"possible number of events}$

#=(2xx5!xx5!)/(10!)=1/126#

One precaution

Calculating the favourable number of events if we consider that Boys or Girls may take positions in 4 gaps in between two and 2 positions at two ends i.e. total 6 possible positions. then we arrive at an erroneous possible number of events as #6!xx5!#.

This is wrong.Because in this calculation 2 Boys or 2 Girls may come side by side in some of the arrangements, which is not desired here.

## How do you answer something like 7c3 for a combination and 8p3 for a permutation?

Parzival S.
Featured 2 months ago

#P_(8,3)=(8!)/((8-3)!)=(8!)/(5!)=(8xx7xx6xx5!)/(5!)=8xx7xx6=336#

#C_(7,3)=(7!)/((3!)(7-3)!)=(7!)/((3!)(4!))=(7xx6xx5xx4!)/(3xx2xx4!)=7xx5=35#

#### Explanation:

Combinations, Permutations, and Factorials are used to help calculate the numbers of ways something can be done or ordered.

For instance, let's say we have 5 distinguishable books and want to arrange them on a shelf. How can we do this?

There are 5 books we can place in the first position, 4 in the second, 3 in the third, etc... and so we can express the number of ways we can order the books on the shelf by saying:

$5 \times 4 \times 3 \times 2 \times 1 = 120$

Because we do this so often in Permutation and Combination problems, it'd be convenient to have some sort of symbol to let us know that we want to multiply all the natural numbers up to a given value - and this is done via the Factorial :

#5xx4xx3xx2xx1=5! = 120#

Now let's say we have 7 books and want to arrange 5 on the shelf. We could use the same sort of logic as before, see that there are 7 books we can put in the first position, 6 in the second, 5 in the third, etc, all the way to 3 at the fifth and final book. I can write that as:

$7 \times 6 \times 5 \times 4 \times 3$

and I can sit down and multiply this out each time depending on the number of books available and the ones I intend to arrange. Or I can see that if I simply do this:

$\frac{7 \times 6 \times 5 \times 4 \times 3 \textcolor{red}{\times 2 \times 1}}{\textcolor{red}{2 \times 1}}$

I can now express this in terms of two factorials:

#(7xx6xx5xx4xx3color(red)(xx2xx1))/color(red)(2xx1)=(7!)/(2!)#

and I can even go one more step and see that I can express the denominator in terms of the number of books I have and the number of books on the shelf:

#(7!)/(2!)=(7!)/((7-5)!)#

And with this we have the general formula for a Permutation :

#P_(n,k)=(n!)/((n-k)!); n="population", k="picks"#

We can work the question:

#P_(8,3)=(8!)/((8-3)!)=(8!)/(5!)=(8xx7xx6xx5!)/(5!)=8xx7xx6=336#

That equation works well if we care which book is in which position on the shelf. But what if we don't care? If we simply say that simply having the History book, the English book, and the Math book on the shelf together in any order counts as 1 way or ordering (this is much like a poker hand - we don't care in what order we draw the cards, we simply want the best hand possible)? To do this, we divide the Permutation by the number of items picked - we are essentially taking out the "repeats". Sticking with the 7 books available and 5 books to be placed on a shelf:

#(7!)/((5!)(2!))=(7!)/(5!(7-5)!)#

And with this we have the general formula for a Combination:

#C_(n,k)=(n!)/((k!)(n-k)!)# with $n = \text{population", k="picks}$

We can work the question:

#C_(7,3)=(7!)/((3!)(7-3)!)=(7!)/((3!)(4!))=(7xx6xx5xx4!)/(3xx2xx4!)=7xx5=35#

## How do you create a box plot for the data set {1, 12, 2, 9, 2, 3, 7, 3, 3, 6)?

Steve
Featured 4 weeks ago

The raw data is:

$\left\{1 , 12 , 2 , 9 , 2 , 3 , 7 , 3 , 3 , 6\right\}$

First we arrange in ascending order:

$\left\{1 , 2 , 2 , 3 , 3 , 3 , 6 , 7 , 9 , 12\right\}$

We the calculate the media, ${Q}_{2}$, which is the middle value, and as we have an even number of data points will be an average:

${Q}_{2} = \frac{1}{2} \left(3 + 3\right) = 3$

Similarly we calculate the lower and upper quartiles:

${Q}_{1} = 2$
${Q}_{3} = 7$

So then we have enough information to draw the box and whisker diagram:

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