The Independent variable is what the experimenters are directly manipulating-- here, that is the strength of the Chemical XYZ.

The Dependent variable is the variable that changes based on the Independent variable-- in this case, it is the amount of oil on participants' scalps. 

A Dependent variable is what changes based on the independent, manipulated variable.

The Independent variables are the Room Temperature and the Lighting state, which the researchers were manipulating themselves.

The Dependent variable is the Mouse Food Intake, which is the variable that changed based on how the independent variables changed.

Remember how the mean and standard deviation of a random variable are affected when it is multiplied by a constant.

$\displaystyle E(cX)=c\cdot E(X)$

$\displaystyle \mu = 2\cdot 10 = 20$

$\displaystyle \sigma=\sqrt{c^2\cdot 5}$

$\displaystyle \sigma = \sqrt{2^2 \cdot 5} = \sqrt{20}$

Any linear combination of independent random variables is also normally distributed with the mean and variance depending on the weights on the random variables. The mean is additive in the sense that

$\displaystyle \small mean\left(\frac{X_1+...+X_{10}}{10}\right)=\frac{1}{10}\left(mean(X_1)+...+mean(X_{10})\right)$

Each $\displaystyle \small mean(X_i)$ is $\displaystyle \small 0$, so the sum is equal to zero.

This means the sum of the average 

$\displaystyle \small \small \small \frac{X_1+X_2+...+X_{10}}{10}$ is $\displaystyle \small 0$.

The variance satisfies

$\displaystyle Var\left(\frac{X_1+X_2+...+X_{10}}{10}\right)=\frac{1}{10^2}Var(X_1+...+X_{10})$

$\displaystyle \small =\frac{1}{100}(Var(X_1)+...Var(X_{10}))=\frac{10}{100}=\frac{1}{10}$ because of independence.

This means that the average is normally distributed with mean $\displaystyle \small 0$ and variance $\displaystyle \small \frac{1}{10}$.

There is a relatively simple way of doing this problem. The sum of any set of independent normal random variables is also distributed normally. So $\displaystyle \small X_1+X_2+X_3$ has a normal distribution. Now we can compute the mean and variance. The mean is additive:

$\displaystyle \small mean(X_1+X_2+X_3)=mean(X_1)+mean(X_2)+mean(X_3)$

$\displaystyle \small \small =\frac{1}{4}-1+\frac{3}{4}=0$

Variance is also additive in some sense, when the random variables are independent:

$\displaystyle \small \small Var(X_1+X_2+X_3)=Var(X_1)+Var(X_2)+Var(X_3)$

$\displaystyle \small =\frac{1}{2}+\frac{3}{8}+\frac{1}{8}=1$

Thus, $\displaystyle \small X_1+X_2+X_3$ is normally distributed with mean $\displaystyle \small 0$ and variance $\displaystyle \small 1$.

This sum is a standard normal distribution.

The chance that $\displaystyle \small X_1+X_2+X_3< -1$ is thus $\displaystyle \small 0.1587$, if we use a normal table.

To find the mean you sum up all of your values then divide by the total amount of values.  The total sum of the weights is $\displaystyle 60.5lbs$ and there are 10 melons.  $\displaystyle 60.5\div10=60.5lbs$

The z-score equation is given as: z = (X - μ) / σ, where X is the value of the element, μ is the mean of the population, and σ is the standard deviation. To solve for the student's test score (X): 

X = ( z * σ) + 95 = ( 1.10 * 12) + 95 = 108.2.

First, find that $\displaystyle 2A$ has $\displaystyle \mu=2\cdot 5 = 10$ and standard deviation $\displaystyle \sigma = 2 \cdot 4 = 8$.

Then find the mean and standard deviation of $\displaystyle 2A + B$.

$\displaystyle \mu = 10 + 9 = 19$

$\displaystyle \sigma = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$

The variance of a discrete random variable is computed as

$\displaystyle \small Var(X)=\sum_x (x-\mathbb{E}[x])^2P(X=x)$

for all the values of $\displaystyle \small x$ that the random variable $\displaystyle \small X$ can take.

First, we compute $\displaystyle \small \mathbb{E}[X]$, which is the expected value. In this case, it is $\displaystyle \small 2.5$.

So we have

$\displaystyle \small \small Var(X)=\frac{1}{4}(1-2.5)^2+\frac{1}{4}(2-2.5)^2+\frac{1}{4}(3-2.5)^2+\frac{1}{4}(4-2.5)^2$

$\displaystyle \small =\frac{1}{4}(2(1.5^2)+2(0.5^2))=\frac{1}{4}(2(2.25)+2(0.25))=\frac{1}{4}(4.5+0.5)$

$\displaystyle \small \small =\frac{5}{4}=1.25$

Note that the weight of a packed box = weight of books + weight of box + weight of packing material used. 

It is given that $\displaystyle \sigma_{books} = 2\: \mbox{pounds}, \sigma_{box} = 0.15\: \mbox{pounds}, \mbox{and}\: \sigma_{plastic packaging} = 0.25\: \mbox{pounds}$.

The calculation of the standard deviation of the weights of the packed boxes is 

$\displaystyle \\ \sigma_{packed boxes} = \sqrt{\sigma_{clothes} ^{2} + \sigma_{clothes} ^{2} + \sigma _{plastic packaging}^{2}} \\ \\ = \sqrt{2^{2}+ 0.15^{2} + 0.25^{2}} \\ = 2.021\: \mbox{pounds}$