$45$ percentage of population of a large community is in favor of a proposed rise in school taxes.
$i)$ Approximate the probability that a random sample of $1000$ people will contain at least $500$ who are in favor of the proposition.
$ii)$ Approximate the probability that the random sample of $1000$ will contain between $600$ and $700$ inclusive who are in favor
MY WORKING:
Let $X$ be the random variable denoting no. of people out of sample of size $n=1000$ then $P(success)=0.45$
So, $X$~$Bin(0.45,1000)$. Since $n=1000$ is large so $X$ can be approximated as normal random variable $Z$ with parameters given by:
$mean=m=np=1000*0.45=450$
$variance=v^{2}=np(1-p)=1000*0.45*0.55=247.5$
Thus,
$X$~$Bin(0.45,1000)\approx N(450,247.5)$
So,
$i)$ $Pr[X\geq500]=1-Pr[\frac{X-m}{v}<\frac{500-450}{15.732}]=1-Pr[Z<-3.17]=1-Pr[Z>3.17]=P[Z<3.17]=0.9992$
$ii)$ $P[500\leq X\leq700]=Pr[\frac{600-450}{15.732}\leq\frac{X-m}{v}\leq\frac{700-450}{15.732}]=Pr[9.534\leq Z\leq15.891]=Pr[Z\leq15.891]-Pr[Z\leq9.534]$
Now I think My answer to Part $i)$ is correct but I need confirmation. But as far as part $ii)$ is concerned the last values of $Pr[Z\leq15.891]$ & $Pr[Z\leq9.534]$ can not be found from normal table and doesn't make any sense. Where am I making the mistake. Kindly Guide me. Thanks
Your answers are almost correct:
First of all you missed the "$1-$" in the $i)$ part. Therefore you should get $0.0008$. As $99.7\%$ of outcomes are falling into $[-3,3]$, it is reasonable to take $P(Z\geq k)=0$ and $P(Z\leq k)=1$ for all $k > 3$. The values you calculated for $k$ are simply too large and the corresponding probabilities too small, so listing them anywhere doesn't make sense. Therefore you will get $0$ at part $ii)$.
These two solutions seem to be very low. However, remember that the $95\%$ confidence interval is just $\pm 2 \sigma$. As $\sigma \approx 15$ this means that the outcome will very likely be in the interval $[420, 480]$.
If you can't believe this outcome, this is what you get if you calculate the exact results using python and
scipy.spatial.binom:binom.cdf(k, n, p)calculates the probability $P(X \leq k)$ for $X\sim \text{Bin}(n,p)$.