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## The mean of 8 numbers is 41 The mean of 2 of the numbers is 29 What is the mean of the other 6 numbers?

Rhys
Featured 5 months ago

Mean = 45

#### Explanation:

Let the 8 numbers be: $a , b , c , d , e , f , g , h$

$\implies \frac{a + b + c + d + e + f + g + h}{8} = 41$

$\implies a + b + c + d + e + f + g + h = 328$

Let the two of those numbers just be $g \mathmr{and} h$

$\implies \frac{g + h}{2} = 29$

$\implies g + h = 58$

So $\implies a + b + c + d + e + f + \textcolor{b l u e}{g + h} = 328$ Becomes...

$\implies a + b + c + d + e + f + \textcolor{b l u e}{58} = 328$

$\implies a + b + c + d + e + f = 270$

Mean of these six numbers...

#=> color(red)((a+b+c+d+e+f )/6 = 270/6 = 45 #

## Four cards are drawn out from a packet of cards casually. What is the probability to find 2 cards of them to be spade? @probability

Parzival S.
Featured 3 months ago

#(57,798)/(270,725)~~21.35%#

#### Explanation:

Let's first see the number of ways we can pick 4 cards from a pack of 52:

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

#C_(52,4)=(52!)/((4!)(48!))=(52xx52xx50xx49)/24=270,725#

How many ways can we draw 4 cards and have exactly 2 of them be spades? We can find that by choosing 2 from the population of 13 spades, then choosing 2 cards from the remaining 39 cards:

#C_(13,2)xxC_(39,2)=(13!)/((2!)(11!))xx(39!)/((2!)(37!))=(13xx12)/2xx(39xx38)/2=57,798#

This means the probability of drawing exactly 2 spades on a 4 card draw from a standard deck is:

#(57,798)/(270,725)~~21.35%#

## A standard number cube is tossed. What is the probability that the number will be less than or equal to 3?

EZ as pi
Featured 3 months ago

$P \left(3 \text{ or less on a die}\right) = \frac{3}{6} = \frac{1}{2}$

#### Explanation:

To determine the probability of an event happening, there are two values you need to determine first.

1. How many possible outcomes are there altogether?
In this case with a toss of the die, there are $6$ outcomes, being the numbers from $1$ to $6$.

2. Of the possible outcomes how many of them will meet our conditions? IN this case we want a number equal to or less than $3$, ... the numbers $1 , 2 , 3$. There are $3$ desired outcomes.

$\text{Probability" = "number of desirable outcomes"/"total number of possible outcomes}$

$P \left(3 \text{ or less on a die}\right) = \frac{3}{6} = \frac{1}{2}$

Write a probability as a fraction and give in its simplest form.

## What is the variance for the following data, 2 4 5 7 ? Please show working.[steps].

Featured 1 month ago

$\textcolor{red}{{\sigma}^{2} = 3.25}$

#### Explanation:

To find the variance, we first need to calculate the mean.

To calculate the mean, simply add all the data points, then divide by the number of data points.

The formula for the mean $\mu$ is

$\mu = \frac{{\sum}_{k = 1}^{n} {x}_{k}}{n} = \frac{{x}_{1} + {x}_{2} + {x}_{3} + \cdots + {x}_{n}}{n}$

Where ${x}_{k}$ is the $k$th data point, and $n$ is the number of data points.

For our data set, we have:

$n = 4$

$\left\{{x}_{1} , {x}_{2} , {x}_{3} , {x}_{4}\right\} = \left\{2 , 4 , 5 , 7\right\}$

So the mean is

$\mu = \frac{2 + 4 + 5 + 7}{4} = \frac{18}{4} = \frac{9}{2} = 4.5$

Now to calculate the variance, we find out how far away each data point is from the mean, then square each of those values, add them up, and divide by the number of data points.

The variance is given the symbol ${\sigma}^{2}$

The formula for the variance is:

${\sigma}^{2} = \frac{{\sum}_{k = 1}^{n} {\left({x}_{k} - \mu\right)}^{2}}{n} = \frac{{\left({x}_{1} - \mu\right)}^{2} + {\left({x}_{2} - \mu\right)}^{2} + \ldots + {\left({x}_{n} - \mu\right)}^{2}}{n}$

So for our data:

${\sigma}^{2} = \frac{{\left(2 - 4.5\right)}^{2} + {\left(4 - 4.5\right)}^{2} + {\left(5 - 4.5\right)}^{2} + {\left(7 - 4.5\right)}^{2}}{4}$

${\sigma}^{2} = \frac{{\left(- 2.5\right)}^{2} + {\left(- 0.5\right)}^{2} + {\left(0.5\right)}^{2} + {\left(2.5\right)}^{2}}{4}$

${\sigma}^{2} = \frac{6.25 + 0.25 + 0.25 + 6.25}{4} = \frac{13}{4} = \textcolor{red}{3.25}$

## I have three distinct mystery novels, three distinct fantasy novels, and three distinct biographies. I'm going on vacation, and I want to take two books of different genres. How many possible pairs can I choose?

Parzival S.
Featured 1 month ago

$\left(\begin{matrix}3 \\ 2\end{matrix}\right) {\left(\begin{matrix}3 \\ 1\end{matrix}\right)}^{2} = 3 \times {3}^{2} = 27$

#### Explanation:

Let's first notice that this is a combination problem because we don't care in what order we pick the books (if we did, that'd be a permutation problem).

The general formula for a combination is:

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

To do this, let's first see that we're choosing 2 categories from the population of 3 categories. That's

$\left(\begin{matrix}3 \\ 2\end{matrix}\right)$

For each of the categories, there are 3 books we can choose from and we want 1. For each category chosen, then, we have:

$\left(\begin{matrix}3 \\ 1\end{matrix}\right)$

Since we have 2 categories chosen, we need to square this:

${\left(\begin{matrix}3 \\ 1\end{matrix}\right)}^{2}$

And so altogether we have:

$\left(\begin{matrix}3 \\ 2\end{matrix}\right) {\left(\begin{matrix}3 \\ 1\end{matrix}\right)}^{2} = 3 \times {3}^{2} = 27$

We can list them out (I'll use A, B, and C; L, M, and N; and W, X, and Y to list out the books):

AL BL CL
AM BM CM
AN BN CN

AW BW CW
AX BX CX
AY BY CY

LW MW NW
LX MX NX
LY MY NY

## Are population means score in statistics different between 2006 and 2016 students? ( Help ) (Stats)

Anjali G
Featured 3 weeks ago

Part a) No, there is not a significant difference between 2006 and 2016 students' test scores.

Part b) Interval: $\left(- 0.689 , 5.089\right)$

#### Explanation:

Part A

We want to determine whether there is a significant difference between the population mean scores of the students. We will test the hypotheses:

${H}_{0} : {\mu}_{\text{y" = mu_"x}}$
${H}_{a} : {\mu}_{\text{y" < mu_"x}}$

where "y" represents data from 2016, and "x" represents data from 2006.

We can conduct a two-sample t-test for the difference between two means using a significance level $\alpha = .05$ if the following conditions for inference are met.

• Random: both samples (2006 and 2016) are random samples
• 10% condition / Independence: We must assume that the population of test takers ...
• in 2006 is greater than 140 ${N}_{x} \ge 10 \cdot 14$
• in 2016 is greater than 200 ${N}_{y} \ge 10 \cdot 20$
• Large/Normal samples: We must assume that the distributions for both populations are each approximately Normal

The formula for the test statistic t is:

$t = \frac{\overline{x} - \overline{y}}{\sqrt{\frac{{s}_{x}^{2}}{{n}_{x}} + \frac{{s}_{y}^{2}}{{n}_{y}}}}$

with degrees of freedom (using the lower n) $\text{df} = {n}_{x} - 1$

Substitute values:
$t = \frac{73.0 - 70.8}{\sqrt{\frac{{3.2}^{2}}{14} + \frac{{4.6}^{2}}{20}}}$

$\text{df} = 14 - 1 = 13$

Now, using the table of t critical values or a calculator, we can find that our $p$ value lies between $.05$ and $.10$.
(Using a calculator):

$p = .0549$

Since our $p$ value $p = .0549$ is greater than our significance level $\alpha = .05$, we fail to reject our null hypothesis. There is not convincing evidence of a significant difference between 2006 and 2016 tests.

Part B

We want to construct a 95% confidence interval for the difference between two means.

We can use a two-sample t-interval . The conditions for inference were verified in part (a).

The formula for the two-sample t-interval with 95% confidence is:

$\left(\overline{x} - \overline{y}\right) \pm {t}^{\text{*}} \sqrt{\frac{{s}_{x}^{2}}{{n}_{x}} + \frac{{s}_{y}^{2}}{{n}_{y}}}$

Substitute values:
Find t using the table of critical t values (linked above). This is where we specify the 95% confidence.

$\left(73.0 - 70.8\right) \pm \left(2.160\right) \sqrt{\frac{{3.2}^{2}}{14} + \frac{{4.6}^{2}}{20}}$

$2.2 \pm 2.889$

We are 95% confident that the interval from $- 0.689$ to $5.089$ captures the true difference $\overline{x} - \overline{y}$ between the population mean test scores in 2006 and 2016.

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