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The survival function of this exponential distribution is $P(T>t)=e^{-\alpha t}$. At the expected value found above,

The variance is the second moment minus the square of the expected value:

The expected value of the discrete uniform distribution is

The density is symmetric about $7/2$, giving expected value $\mathbb{E}[X]=7/2$. Its second moment is $\mathbb{E}[X^2]=\frac{1}{4}({\int }_1^3{x}^{2}dx+{\int }_4^6{x}^{2}dx)=89/6$. ThereforeEquivalently, each equally weighted interval has variance $1/3$ and its centre is $3/2$ from the overall mean, giving $1/3+9/4$ by the law of total variance.

Write the chocolate count as $K_n={\sum }_{j=1}^n{I}_j$ with independent Bernoulli random variables. Each has expected value $p$ and variance $p(1-p)$. By linearity of expected value and additivity of variance for independent variables,

For a continuous random variable with probability density function $p$, its expected value and variance areThe mean requires an integrable first absolute moment, and a finite variance requires a finite second moment. When these hold, $\mathrm{Var}X=\mathbb{E}[X^2]-(\mathbb{E}X{)}^2$.

The lazy choice weights are $1/w$ and $1/(3w)$, in ratio three to one. Normalizing gives $\boxed{l=3/4}$. Let $Q$ denote selection of the first question and let $q=P(\text{Lazy}\mid Q)$. Its conditional expected value, divided by $m$, isso $q=6/7$. If the overall lazy fraction is $p$, the law of total probability gives $P(Q)=(3/4)p+(1/2)(1-p)=(p+2)/4$. Bayes' theorem then givesSolving $6(p+2)=21p$ yields $\boxed{p=4/5}$. The observed mean is conditioned on the chosen question, so it cannot be equated directly to the unconditional mean over all students.

Write the count as a sum of $24$ Bernoulli random variables, one for each attendance indicator. Each indicator random variable has expected value $3/4$, so linearity of expectation givesEquivalently, $N$ has a binomial distribution with parameters $24$ and $3/4$.

The expected value is the improper integral $\alpha {\int }_1^{\infty }{x}^{-\alpha }dx$. It converges exactly when $\alpha >1$, yieldingAt $\alpha =1$ this improper integral diverges logarithmically, and for $0<\alpha <1$ it diverges as a power of its upper endpoint. In both cases the nonnegative variable has infinite mean. This is the first Pareto moment threshold.

Write $p_j=\mathbb{P}(X=j)$, with ${\sum }_{j=0}^N{p}_j=1$. The expected value, variance, and strict lower-tail probability areThe upper summation limit $Y-1$ is essential because the inequality is strict and the variable is integer-valued.

The three preceding probabilities satisfyTheir expected value and second moment areSubtracting the squared mean givesThe value $\mu =2(2/10)$ also follows from linearity of expectation. The variance includes the finite-population reduction caused by sampling without replacement.

Normalize the Rayleigh distribution by substituting $s=x^2/(2{\sigma }^2)$:For its sketch, the density starts at zero, is positive for $x>0$, and tends to zero at infinity. Differentiation givesIt therefore has a single maximum at $x=\sigma$, with height $e^{-1/2}/\sigma$.

The expected value isIntegrating by parts with $xdx{e}^{-x^2/(2{\sigma }^2)}=-{\sigma }^{2}d(e^{-x^2/(2{\sigma }^2)})$ makes the boundary term vanish and leaves ${\int }_0^{\infty }{e}^{-x^2/(2{\sigma }^2)}dx$. The Gaussian integral gives

The second moment isSubtracting the squared expected value gives the variance and standard deviation:The variance is nonnegative because $0<r\le R$. Both $\mu$ and $\sigma$ have units of time. At $r=R$ the result reduces to $R/\sqrt{12}$, the standard deviation of a uniform distribution over one full cycle.

The atom at zero contributes zero to the expected value, while all positive delays must retain their unconditional density. ThereforeThis mean is $r/R$ times the mean $r/2$ conditional on encountering a stop; using $r/2$ alone would omit the vehicles passing without delay.

Compute the first two moments from the probability distribution:The variance is $E[X^2]-E[X{]}^2=9/5-36/25=9/25$, so the expected value and standard deviation are

The same integration by substitution gives the expected valueThe remaining integrand is an odd function. Its integral is absolutely convergent, since ${\mathrm{sech}}^{2}t\le 4e^{-2|t|}$ and $|t|e^{-2|t|}$ is integrable. Thus cancellation under symmetry in integration is legitimate, rather than merely a principal-value calculation. ConsequentlyEquivalently, the logistic distribution is symmetric about $\mu$, and its exponentially decaying tails ensure that its expected value exists.

Write $P_k=P(X=k)$ for $k=0,1,2,\dots$, with ${\sum }_k{P}_k=1$. Provided the relevant moments exist, the expected value and variance areThe displayed support includes zero, so these are nonnegative integer values.

The zero-goal term contributes nothing to the expected value. Shift the summation index in the remaining terms:The exponential series justifies the convergent reindexing. Thus the parameter is the mean goals per game.

The expected value is not defined: the integral of $yf(y)$ over the positive half-line diverges logarithmically to $+\infty$, and the negative half-line gives $-\infty$. In particular, symmetry alone does not supply a mean. More explicitly,The second moment also diverges because $y^{2}f(y)\to x_0/\pi >0$ as $|y|\to \infty$. ThusThe symmetric Cauchy principal value of the first-moment integral is zero, but it is not an expected value and does not define a variance.

The new angular interval has width $\pi /2$, so its probability density function is $2/\pi$ there. Its endpoints stay away from the poles of the tangent function, making $Y$ bounded and its expected value finite. Integrate over the continuous uniform distribution of angles:The two cosines are $\sqrt{3}/2$ and $1/2$. HenceIt has units of length and is positive because the permitted positive angles extend farther than the negative ones.