Solution. (a). The interval is \(4200\pm 600\), and \(600=k\sigma \) with \(k=\frac {600}{300}=2\). Chebyshev’s inequality gives \[P\left (\left |X-4200\right |\geq 600\right )\leq \frac {1}{2^2}=0.25,\] \[\therefore \quad P\left (3600<X<4800\right )\geq 1-0.25=0.75.\]

(b). If \(X\sim N\left (4200,300^2\right )\) then \[P\left (3600<X<4800\right )=P(-2<Z<2)=0.9772-0.0228=0.9544.\]

Solution. Write \(q=1-p\) and \(\sigma _n=\sqrt {npq}\). Since \(M_{Z_n}(t)=\left (q+pe^t\right )^n\), \[M_{Y_n}(t)=e^{-\frac {npt}{\sigma _n}} \left (q+pe^{\frac {t}{\sigma _n}}\right )^n,\qquad \text {so}\qquad \ln M_{Y_n}(t)=-\frac {npt}{\sigma _n} +n\ln \left (q+pe^{\frac {t}{\sigma _n}}\right ).\] Expand the exponential, writing \(s=\frac {t}{\sigma _n}\to 0\): \[q+pe^{s}=q+p\left (1+s+\frac {s^2}{2}+\cdots \right ) =1+ps+\frac {ps^2}{2}+\cdots ,\] since \(q+p=1\). Now use \(\ln (1+w)=w-\frac {w^2}{2}+\cdots \) with \(w=ps+\frac {ps^2}{2}+\cdots \): \[\ln \left (q+pe^{s}\right )=ps+\frac {ps^2}{2}-\frac {p^2s^2}{2}+O\left (s^3\right ) =ps+\frac {pq\,s^2}{2}+O\left (s^3\right ),\] because \(\frac {p}{2}-\frac {p^2}{2}=\frac {p(1-p)}{2}=\frac {pq}{2}\). Therefore \[\ln M_{Y_n}(t)=-\frac {npt}{\sigma _n}+n\left [\frac {pt}{\sigma _n} +\frac {pq\,t^2}{2\sigma _n^2}+O\!\left (\frac {1}{n^{\frac 32}}\right )\right ] =\frac {npq\,t^2}{2npq}+O\!\left (\frac {1}{\sqrt n}\right ) \ \longrightarrow \ \frac {t^2}{2}.\] \[\therefore \quad M_{Y_n}(t)\longrightarrow e^{\frac {t^2}{2}}, \qquad \text {so}\qquad \frac {Z_n-np}{\sqrt {np(1-p)}}\ \overset {d}{\longrightarrow }\ N(0,1).\]

Solution. (a). For \(\text {POI}(1)\) both the mean and the variance are \(1\), so \(\mu =1\), \(\sigma =1\), and \(Y_n=\sqrt n\left (\bar X-1\right )\) immediately.

A sum of independent Poisson variables is Poisson with the means added, so \(S=\sum _{i=1}^nX_i\sim \text {POI}(n)\) and \(M_S(t)=e^{n\left (e^t-1\right )}\). Now \[Y_n=\sqrt n\left (\frac {S}{n}-1\right )=\frac {S-n}{\sqrt n},\] so \[M_{Y_n}(t)=E\left [e^{\frac {t(S-n)}{\sqrt n}}\right ] =e^{-t\sqrt n}\,M_S\!\left (\frac {t}{\sqrt n}\right ) =\exp \left [-t\sqrt n+n\left (e^{\frac {t}{\sqrt n}}-1\right )\right ]. \qquad \blacksquare \]

(b). This is exactly the moment generating function computed in the worked example of section 1.4, where \(Z_n\sim \text {POI}(n)\) and \(Y_n=\frac {Z_n-n}{\sqrt n}\) — and indeed \(S\) here is such a \(Z_n\). Expanding as there, \[\ln M_{Y_n}(t)=\frac {t^2}{2}+O\!\left (\frac {1}{\sqrt n}\right ) \ \longrightarrow \ \frac {t^2}{2},\] \[\therefore \quad Y_n\ \overset {d}{\longrightarrow }\ N(0,1).\]

Solution. First the population mean and variance: \[\mu =\int _0^12x^2dx=\frac 23,\qquad E\left (X^2\right )=\int _0^12x^3dx=\frac 12,\] \[\sigma ^2=\frac 12-\left (\frac 23\right )^2=\frac 12-\frac 49=\frac {1}{18}.\] So \(\bar X\) has mean \(\frac 23=0.6667\) and variance \(\frac {1}{18\times 20}=\frac {1}{360}\), giving \[\sigma _{\bar X}=\frac {1}{\sqrt {360}}=0.0527.\] By the central limit theorem \(\bar X\) is approximately normal, so \[P\left (0.60<\bar X<0.75\right ) \approx P\!\left (\frac {0.60-0.6667}{0.0527}<Z<\frac {0.75-0.6667}{0.0527}\right ) =P(-1.26<Z<1.58),\] \[P\left (0.60<\bar X<0.75\right )\approx 0.9429-0.1038=0.8391.\]

Solution. Here \(\mu =np=45\) and \(\sigma ^2=np(1-p)=150(0.3)(0.7)=31.5\), so \(\sigma =5.6125\).

(a). The event \(Y\geq 50\) becomes \(Y>49.5\): \[P(Y\geq 50)\approx P\!\left (Z>\frac {49.5-45}{5.6125}\right )=P(Z>0.80) =1-0.7881=0.2119.\] The exact binomial value is \(0.2101\).

(b). \[P(Y=45)\approx P\left (44.5<Y<45.5\right ) =P\!\left (\frac {-0.5}{5.6125}<Z<\frac {0.5}{5.6125}\right )=P(-0.089<Z<0.089),\] \[P(Y=45)\approx 0.5355-0.4645=0.0710,\] against an exact value of \(0.0709\).

Solution. A \(\chi ^2(k)\) distribution has mean \(k\) and variance \(2k\), so here \(\mu =20\) and \(\sigma ^2=40\). Then \[\text {Var}\left (\bar X\right )=\frac {40}{50}=0.8,\qquad \sigma _{\bar X}=\sqrt {0.8}=0.8944.\] By the central limit theorem, \[P\left (19<\bar X<21\right ) \approx P\!\left (\frac {19-20}{0.8944}<Z<\frac {21-20}{0.8944}\right ) =P(-1.12<Z<1.12),\] \[P\left (19<\bar X<21\right )\approx 0.8686-0.1314=0.7372.\]

Solution. For \(\text {GAM}(\alpha ,\beta )\) with \(\alpha =2\) and \(\beta =3\), \[\mu =\alpha \beta =6,\qquad \sigma ^2=\alpha \beta ^2=18.\] So \(\text {Var}\left (\bar X\right )=\frac {18}{64}=0.2813\) and \(\sigma _{\bar X}=0.5303\). Then \[P\left (5<\bar X<7\right ) \approx P\!\left (\frac {5-6}{0.5303}<Z<\frac {7-6}{0.5303}\right ) =P(-1.89<Z<1.89),\] \[P\left (5<\bar X<7\right )\approx 0.9706-0.0294=0.9412.\]

Solution. The two sample means are normal and independent: \[\bar X\sim N\!\left (10,\frac {25}{36}\right ),\qquad \bar Y\sim N\!\left (8,\frac {16}{36}\right ).\] A difference of independent normal variables is normal, with the means subtracted and the variances added: \[\bar X-\bar Y\sim N\!\left (2,\ \frac {25}{36}+\frac {16}{36}\right ) =N\!\left (2,\ \frac {41}{36}\right ),\qquad \text {sd}=\sqrt {\frac {41}{36}}=1.0672.\] Therefore \[P\left (\bar X>\bar Y\right )=P\left (\bar X-\bar Y>0\right ) =P\!\left (Z>\frac {0-2}{1.0672}\right )=P(Z>-1.87)=0.9693.\]