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## What does a high/low standard deviation mean in real terms?

Kate M.
Featured 4 months ago

The higher the standard deviation the more variability or spread you have in your data.

#### Explanation:

Standard deviation measures how much your entire data set differs from the mean.

The larger your standard deviation, the more spread or variation in your data. Small standard deviations mean that most of your data is clustered around the mean.

In the following graph, the mean is 84.47, the standard deviation is 6.92 and the distribution looks like this:

Many of the test scores are around the average. There's one student who scored a 96, two students who scored 69, another two who scored 71, but most students scored close to somewhat close to the average of 84.47.

In this second graph, the mean is 80, the standard deviation is 14.57 , and the distribution looks like this:

There is greater variability in the test scores. One student scored a 24, which is pretty far from the average test score of 80. Another student scored a 45, which also isn't close to 80.

## Two balls are drawn from a bag containing 5 white and 7 black balls at random. What is the probability that they would be of different colours?

EZ as pi
Featured 3 months ago

If the first ball is replaced: $P \left(\text{different}\right) = \frac{35}{72}$

If the first ball is not replaced: $P \left(\text{different}\right) = \frac{35}{66}$

#### Explanation:

There are $3$ outcomes that can occur:

• both are white
• both are black
• one of each color

The probability of one of each color is $1 - P \left(\text{same}\right)$

If the first ball is replaced:

$P \left(W W\right) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}$
$P \left(B B\right) = \frac{7}{12} \times \frac{7}{12} = \frac{49}{144}$

#:. P("same) = 25/144+49/144 = 74/144#

$P \left(\text{different}\right) = 1 - \frac{74}{144} = \frac{70}{144} = \frac{35}{72}$

However, if the first ball is NOT replaced, the probability changes for the second ball. There are fewer of one color and one ball less.

$P \left(W W\right) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}$

$P \left(B B\right) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}$

#:. P("same) = 20/132+42/132 = 62/132#

$P \left(\text{different}\right) = 1 - \frac{62}{132} = \frac{70}{132} = \frac{35}{66}$

## Help me answer this question.....? (question in picture below)

Geoff K.
Featured 3 months ago

(i) $c = \frac{1}{10} , \text{ " d=1/6" }$ (ii) $\text{Var} \left(X\right) = \frac{16}{3}$

(iii) $\text{P"(X_1+X_2=10)=11/100" }$ (iv) $\text{E} \left(Y\right) = 240 ,$

$\text{Var"(Y)=256" }$ (v) $\text{P} \left(220 \le Y \le 260\right) \approx 0.7888$

#### Explanation:

(i) $\text{P"(X < 6) = "P} \left(X \ge 6\right)$

and by total probability

$\text{P"(X < 6) + "P} \left(X \ge 6\right) = 1$

so

$\text{P"(X < 6) = 1/2 = "P} \left(X \ge 6\right)$

Since there are 5 outcomes less than 6, and all 5 outcomes have the same probability, we divide that $\frac{1}{2}$ into 5 equal pieces to give to each of the 5 outcomes: $c = \left(\frac{1}{5}\right) \left(\frac{1}{2}\right) = \frac{1}{10}$.

Similarly, the outcomes 6, 7, and 8 are equally likely, and their probabilities sum to (the other) $\frac{1}{2}$, so $d = \left(\frac{1}{3}\right) \left(\frac{1}{2}\right) = \frac{1}{6.}$

(ii) #"E"(X)=sum_(x=1)^8[x * "P"(X=x)]#

$= {\sum}_{x = 1}^{5} \left[x \cdot \text{P"(X=x)]+sum_(x=6)^8[x * "P} \left(X = x\right)\right]$
$= \frac{1}{10} {\sum}_{x = 1}^{5} x + \frac{1}{6} {\sum}_{x = 6}^{8} x$

$= \frac{15}{10} + \frac{21}{6}$

$= \frac{45 + 105}{30} = \frac{150}{30} = 5$

#"Var"(X)="E"(X^2)-{"E"(X)}^2#

$= {\sum}_{x = 1}^{8} \left[{x}^{2} \text{P} \left(X = x\right)\right] - {5}^{2}$
$= \frac{1}{10} {\sum}_{x = 1}^{5} {x}^{2} + \frac{1}{6} {\sum}_{x = 6}^{8} {x}^{2} - 25$

$= \frac{55}{10} + \frac{149}{6} - 25$

$= \frac{165 + 745 - 750}{30} = \frac{160}{30} = \frac{16}{3}$

(iii) Let ${X}_{1}$ and ${X}_{2}$ be the scores for the first roll and second roll, respectively. Then

$\text{P} \left({X}_{1} + {X}_{2} = 10\right)$

$= \left(\left(\text{P"(X_1=2, X_2=8)),(+"P"(X_1=3, X_2=7)),(+"P"(X_1=4, X_2=6)))+"P"(X_1=5, X_2=5)+(("P"(X_1=6, X_2=4)),(+"P"(X_1=7, X_2=3)),(+"P} \left({X}_{1} = 8 , {X}_{2} = 2\right)\right)\right)$

SInce all 6 terms in the big brackets have the same probability (due to independence of ${X}_{1}$ and ${X}_{2}$), we have

$= 6 \left[\text{P"(X_1=2)"P"(X_2=8)] + ["P"(X_1=5)"P} \left({X}_{2} = 5\right)\right]$

$= 6 \left[\frac{1}{10} \times \frac{1}{6}\right] + \left[\frac{1}{10} \times \frac{1}{10}\right]$

$= \frac{1}{10} + \frac{1}{100}$

$= \frac{10 + 1}{100} = \frac{11}{100}$

(iv) $Y = {\sum}_{i = 1}^{48} {X}_{i}$, where the ${X}_{i} \text{'s}$ are independent. So

$\text{E"(Y)="E} \left({\sum}_{i = 1}^{48} {X}_{i}\right)$
#color(white)("E"(Y))=sum_(i=1)^48 "E"(X_i)" "# (by independence)
$\textcolor{w h i t e}{\text{E} \left(Y\right)} = {\sum}_{i = 1}^{48} 5$

$\textcolor{w h i t e}{\text{E} \left(Y\right)} = 48 \times 5 = 240$

Similarly,

$\text{Var"(Y)="Var} \left({\sum}_{i = 1}^{48} {X}_{i}\right)$
#color(white)("Var"(Y))=sum_(i=1)^48 "Var"(X_i)" "# (by independence)
$\textcolor{w h i t e}{\text{Var} \left(Y\right)} = {\sum}_{i = 1}^{48} \frac{16}{3}$

$\textcolor{w h i t e}{\text{Var} \left(Y\right)} = 48 \times \frac{16}{3} = 256$

(v) Assuming $Y$ is normally distributed, we have

$Y \text{Â ~Â " "N} \left(\mu = 240 , {\sigma}^{2} = 256\right)$

So

$\text{P} \left(220 \le Y \le 260\right)$

$= \text{P} \left(\frac{220 - \mu}{\sigma} \le \frac{Y - \mu}{\sigma} \le \frac{260 - \mu}{\sigma}\right)$

Using the standard normal $\left(Z\right)$ equivalence, this is

$= \text{P} \left(\frac{220 - 240}{16} \le Z \le \frac{260 - 240}{16}\right)$

#="P"("â€“"1.25 <= Z <= 1.25)#

#="P"(Z <= 1.25) - "P"(Z < "â€“"1.25)#

$= \Phi \left(1.25\right) - \Phi \left(\text{â€“} 1.25\right)$

These values can be found by lookup in a $z$-table (or using software). By table lookup:

$= 0.8944 - 0.1056 \text{ } = 0.7888$

## What does it mean for a property to be quantized?

BillytheKid
Featured 3 months ago

It means it can only have discrete amounts...

#### Explanation:

If you drive a car on the motorway and think you are driving exactly 50 mph, chances are you are not: it depends on a myriad of factors.

For instance, how far you have the accelerator pushed down, the friction-coefficient with the road surface (which can vary), speed, strength and angle of a headwind, ambient temperature, vertical gradient (angle) of the road, etc.

Factors like these will change constantly, and even though you try to compensate for them with the accelerator, the result is that your speed will continuously vary and can have ANY value.

In contrast, consider a modern Hi-Fi amplifier/radio/TV.

In virtually all modern sets, when you adjust the volume, that will be done in digital steps: if the volume is set to 15 and you want to increase it to 20, you press the button (often on a remote) and it goes up by increments of 1: 15, 16, 17, 18, 19, and finally 20. No matter how you try, you can NOT adjust it to, let's say, 18.5 or 19.4.

You can consider the steps as Quantised, as they come in little "packets", or "Quantums," of 1 unit. You can't break these down into smaller units.

The term is mostly associated with the atomic model according to Bohr : He had a problem with the model as proposed by Ernest Rutherford.:

According to the Rutherford model, the electron in the Hydrogen atom was circling round the proton. Bohr concluded that it was acting as an oscillating dipole. This would result in loss of Energy, meaning that in the long run the electron would slow down and ultimately crash onto the nucleus (proton).

As this obviously didn't happen, Bohr therefore concluded that the electron couldn't slow down in minute amounts like the car mentioned above can, its Energy could ONLY be acquired or lost in discrete amounts, or "Quantums". Pretty much like you can only adjust the volume of the radio mentioned above in discrete steps....

## What is the difference between discrete probability distribution and continuous probability distribution?

Morgan
Featured 1 month ago

See below.

#### Explanation:

A random variable is a real-valued function defined over a sample space. Consequently, a random variable can be used to identify numerical events of interest in an experiment. Random variables may be either continuous or discrete.

A random variable $Y$ is said to be discrete if it can assume only a finite or countably infinite* number of distinct values.

A set of elements is said to be countably infinite* if the elements in the set can be put into one-to-one correspondence with the positive integers.

Because certain types of random variables occur so frequently in practice, it is useful to have at hand the probability of each value of ra random variable. This collection of probabilities is called the probability distribution of the discrete random variable.

• Distribution functions for discrete random variables are always step functions

Example: Binomial distribution function, $n = 2 , \text{ } p = 1 / 2$

On the other hand, a random variable $Y$ is said to be continuous if it can take on any value in an interval. More precisely, a random variable $Y$ with distribution function $F \left(y\right)$ is said to be continuous if $F \left(y\right)$ is continuous for $- \infty < y < \infty$.

Unfortunately, the probability distribution for a continuous random variable cannot be specified in the same way as outlined above for a discrete random variable; it is mathematically impossible to assign nonzero probabilities to all points on a line interval while satisfying the requirement that the probabilities of the distinct possible values sum to one.

Rather, we define a probability density function for the random variable:

Let $F \left(y\right)$ be the distribution function for a continuous random variable $Y$. Then $f \left(y\right)$, given by

$f \left(y\right) = \frac{\mathrm{dF} \left(y\right)}{\mathrm{dy}} = F ' \left(y\right)$

wherever the derivative exists, is called the probability density function for the random variable $Y$.

Example: A distribution function $F \left(y\right)$ for a continuous random variable

## In a group of 50 persons,30 like tea,25 like coffee and 16 like both.How many like 1)either tea or coffee 2)neither tea or coffee? please represent in cardinal form of set and a venn diagram

tara t.
Featured 4 weeks ago

Either: 23
Neither: 11

#### Explanation:

Using a Venn Diagram - which to me is just a diagrammatic method of representing the information. I just go step by step with whatever information the question gives me, draw it out and then proceed from there.

So from the question, I can see that there are to be two groups of people (those who like tea and those who like coffee). Also, I can gather that there is an overlap in these two groups (i.e. those who like both). Which means that the Venn Diagram looks like this:

Now, the number of people belonging to the red section , the overlap, is information that is given to me in clear words in the question: 16 . The number of people who like both tea and coffee.

This means that now I can, through subtraction, find the number of people who like ONLY tea or ONLY coffee.

ONLY tea: 30 - 16 = 14
ONLY coffee: 25 - 16 = 9

These numbers I would fill in the either of the white sections of the circles.

To find out how many people like neither tea nor coffee, I would add up the three numbers that I have written within the circle and subtract from 50.

9 + 14 + 16 = 39
50 - 39 = 11.

This number I would write outside both circles (as seen in diagram).

The number of people who like EITHER tea or coffee would be the number of people not in the overlap and not outside either circle.

Hence it would be:
14 + 9 = 23

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