Proof. Take the continuous case; the discrete case replaces every integral by a sum. By definition \[\sigma ^2=\int _{-\infty }^{\infty }(x-\mu )^2f(x)\,dx \geq \int _{\left |x-\mu \right |\geq \varepsilon }(x-\mu )^2f(x)\,dx,\] because the discarded region contributes a non-negative amount. On the region that remains, \((x-\mu )^2\geq \varepsilon ^2\), so \[\sigma ^2\geq \varepsilon ^2\int _{\left |x-\mu \right |\geq \varepsilon }f(x)\,dx =\varepsilon ^2\,P\left (\left |X-\mu \right |\geq \varepsilon \right ).\] Dividing by \(\varepsilon ^2\) gives the result. □

Solution. The interval is \(\mu \pm 20\), that is \(\mu \pm k\sigma \) with \(k=\frac {20}{8}=2.5\). Chebyshev gives \[P\left (\left |X-500\right |\geq 20\right )\leq \frac {1}{2.5^2}=0.16,\] \[\therefore \quad P\left (480<X<520\right )\geq 1-0.16=0.84.\] At least \(84\%\) of items are within the stated range — guaranteed, whatever the shape of the distribution. If the weights were known to be normal the answer would be \(0.9876\), but that would be an extra assumption, and the whole value of the bound here is that it needs none.

Proof. \(E\left (\bar X_n\right )=\mu \) and \(\text {Var}\left (\bar X_n\right )=\frac {\sigma ^2}{n}\). By Chebyshev, for any \(\varepsilon >0\), \[P\left (\left |\bar X_n-\mu \right |\geq \varepsilon \right ) \leq \frac {\sigma ^2}{n\varepsilon ^2}\ \longrightarrow \ 0 \quad \text {as }n\to \infty .\] □

Solution. A Poisson variable with mean \(n\) has \(M_{Z_n}(t)=e^{n\left (e^t-1\right )}\), so \[M_{Y_n}(t)=E\left [e^{t\frac {Z_n-n}{\sqrt n}}\right ] =e^{-t\sqrt n}\,M_{Z_n}\!\left (\frac {t}{\sqrt n}\right ) =\exp \left [-t\sqrt n+n\left (e^{\frac {t}{\sqrt n}}-1\right )\right ].\] Expand the exponential inside: \[e^{\frac {t}{\sqrt n}}-1=\frac {t}{\sqrt n}+\frac {t^2}{2n} +\frac {t^3}{6n^{\frac 32}}+\cdots ,\] so \[n\left (e^{\frac {t}{\sqrt n}}-1\right )=t\sqrt n+\frac {t^2}{2} +\frac {t^3}{6\sqrt n}+\cdots \] The \(t\sqrt n\) cancels against \(-t\sqrt n\), and every remaining term after \(\frac {t^2}{2}\) carries a power of \(\frac {1}{\sqrt n}\): \[\ln M_{Y_n}(t)=\frac {t^2}{2}+O\!\left (\frac {1}{\sqrt n}\right ) \ \longrightarrow \ \frac {t^2}{2},\] \[\therefore \quad M_{Y_n}(t)\longrightarrow e^{\frac {t^2}{2}},\] which is the moment generating function of the standard normal. Hence \[\frac {Z_n-n}{\sqrt n}\ \overset {d}{\longrightarrow }\ N(0,1).\]

Sketch. Write \(W_i=\frac {X_i-\mu }{\sigma }\), so the \(W_i\) are independent with mean \(0\) and variance \(1\), and \(Z_n=\frac {1}{\sqrt n}\sum W_i\). If \(W\) has moment generating function \(m(t)\) then \(m(0)=1\), \(m'(0)=0\) and \(m''(0)=1\), so \[m(t)=1+\frac {t^2}{2}+\varphi (t),\qquad \frac {\varphi (t)}{t^2}\to 0\ \text { as }t\to 0.\] By independence, \[M_{Z_n}(t)=\left [m\!\left (\frac {t}{\sqrt n}\right )\right ]^n =\left [1+\frac {t^2}{2n}+\varphi \!\left (\frac {t}{\sqrt n}\right )\right ]^n,\] and the remainder is of smaller order than \(\frac 1n\), so the second limit quoted above applies: \[M_{Z_n}(t)\longrightarrow e^{\frac {t^2}{2}}.\] □

Solution. Here \(\mu =np=32\) and \(\sigma ^2=np(1-p)=80(0.4)(0.6)=19.2\), so \(\sigma =4.3818\).

(a). With the continuity correction, \[P(Y=32)\approx P\left (31.5<Y<32.5\right ) =P\left (\frac {-0.5}{4.3818}<Z<\frac {0.5}{4.3818}\right ) =P(-0.114<Z<0.114),\] \[P(Y=32)\approx 0.5454-0.4546=0.0908,\] against an exact binomial probability of \(0.0908\).

Keep the third decimal place in \(z\) here. Rounding to \(z=\pm 0.11\) gives \(0.0876\), which is out by nearly \(4\%\) — the interval is so narrow that its width, and so the answer, is proportional to \(z\) itself.

(b). \[P(Y\leq 30)\approx P\left (Y<30.5\right ) =P\left (Z<\frac {30.5-32}{4.3818}\right )=P(Z<-0.34)=0.3669,\] against an exact value of \(0.3687\).