edit

64.tex

@@ -28,8 +28,9 @@ P\{X<2\}=p(0)+p(1)&={10!\over0!10!}\times(0.2)^0\times(1-0.2)^{10}
 &=0.3758
 }$$
 
-\bigskip
-\bigskip
+\vfill
+\eject
+
 2. In a large population, 80\% of the members are right-handed.
 
 \bigskip
@@ -139,5 +140,52 @@ i) $P\{0<Z<c\}=0.3531$, $P\{Z<c\}=0.5+0.3531=0.8531$, $c=1.05$
 \bigskip
 j) $P\{-0.54<Z<c\}=0.646$, $P\{Z<c\}=P\{Z<-0.54\}+0.646=0.9406$, $c=1.56$
 
+\vfill
+\eject
+
+5. Let $X$ be a normally distributed random variable with mean 550 and
+standard deviation 150.
+
+\bigskip
+a) $P\{X<790\}=P\{Z<(790-550)/150\}=P\{Z<1.6\}=0.9452$
+
+\bigskip
+b) $P\{220<X<784\}=P\{(220-550)/150<Z<(784-550)/150\}
+=P\{Z<1.56\}-P\{Z<-2.2\}=0.9406-0.0139=0.9267$
+
+\bigskip
+c) $P\{X>211\}=P\{Z>(211-550)/150\}=P\{Z>-2.26\}=P\{Z<2.26\}=0.9881$
+
+\bigskip
+d) $P\{X>c\}=0.1075, (c-550)/150=1.24, c=736$
+
+\bigskip
+e) $P\{X<c\}=0.2005, (c-550)/150=-0.84, c=424$
+
+\vfill
+\eject
+6. Let $X_i$ be independent identically distributed as $N(100,16)$
+for $i=1,2,\ldots,n$.
+
+\bigskip
+a) for $n=25$, what is $P\{\bar X>101.28\}$?
+\medskip
+From $N(100,16)$ we have the mean $\mu=100$ and the
+standard deviation $\sigma=\sqrt{16}=4$.
+By the Central Limit Theorem we have $\sigma_{\bar X}=\sigma/\sqrt n$.
+$$P\{\bar X>101.28\}
+=P\{Z>\sqrt{25}\,(101.28-100)/4\}
+=P\{Z>1.6\}=P\{Z<-1.6\}=0.0548$$
+
+\bigskip
+b) for $n=64$, what is $P\{99<\bar X<101.5\}$?
+$$\eqalign{
+P\{99<\bar X<101.5\}
+&=P\{\sqrt{64}\,(99-100)/4<Z<\sqrt{64}\,(101.5-100)/4\}\cr
+&=P\{-2<Z<3\}\cr
+&=P\{Z<3\}-P\{Z<-2\}\cr
+&=0.9987-0.0228\cr
+&=0.9759
+}$$
 
 \end