Question

Scores for a specialist teacher exam are normally distributed, with a mean of 75 and standard deviation of 5.9
What is the probability that the person scored more than 80 on the specialist teacher exam?
What is the probability that the person scored between 72 and 82 on the specialist teacher exam?
If 38 persons say this specialist teacher exam how many persons are expected to get a score between 72 and 82?
To be eligible for employment as a specialist teacher, you must score in the top 25% what is the lowest score you can earn and still be eligible for employment

B) An average of 3 cars is stolen in the parking lot of a popular mall every month. Find the variance and standard deviation for the number of cars stolen per month
The probability that atleast 5 car thefts will occur in the parking lot during a two month period of November 1st to December 31st 2015

Answer 1

$1) P\{\xi>80\}=1-P\{\xi\leq 80\}=1-F(80)=\\ =1-\frac{1}{\sqrt{2\pi}(5.9)}\int_{-\infty}^{80}e^{-\frac{(x-75)^2}{2(5.9)^2}}dx=1-\Phi(\frac{80-75}{5.9})\approx\\ \approx1-\Phi(0.847)\approx 1-0.8015=0.1985\text{ where }\\ \Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-\frac{t^2}{2}}dx.\\ 2) P\{72<\xi<82\}=F(82)-F(72)=\Phi(\frac{82-75}{5.9})-\Phi(\frac{72-75}{5.9})\approx\\ \approx\Phi(1.186)-\Phi(-0.508)\approx 0.8822-0.3057=0.5765.\\ 3) (0.5765)\cdot 38=21.907.\\ \text{Here we used the fact that }\mu_{\overline{p}}=p.\\ \text{The answer is}\approx 21.\\ 4) F(x)=\frac{1}{\sqrt{2\pi}(5.9)}\int_{-\infty}^{x}e^{-\frac{(t-75)^2}{2(5.9)^2}}dt\\ F(x)=1-0.25=0.75.\\ \text{We should find x}.\\ \text{Substitution: } z=\frac{t-75}{5.9}\\ \Phi(0.75)=0.7734\\ \frac{x-75}{5.9}=0.7734\\ x\approx 79.56.\\ B)\text{Let } \xi \text{ is number of stolen cars during some period of time}.\\ \text{Then }\xi \in \text{Poisson }(\lambda)\text{ where }\lambda \text{ is rate}.\\ \text{We have } \xi \in \text{Poisson }(3), n=1.\\ \text{Hence }D_\xi=\lambda=3, \sigma_\xi=\sqrt{D_\xi}=\sqrt{3}\approx 1.73.\\$

The probability that at least 5 car thefts will occur during a two month period

We have $\lambda=6, n=2.$

$P=1-(P\{\xi=0\}+P\{\xi=1\}+P\{\xi=2\}+P\{\xi=3\}+P\{\xi=4\}).\\ P\{\xi=0\}=\frac{6^0}{0!}e^{-6}=e^{-6}.\\ P\{\xi=1\}=\frac{6^1}{1!}e^{-6}=6e^{-6}.\\ P\{\xi=2\}=\frac{6^2}{2!}e^{-6}=18e^{-6}.\\ P\{\xi=3\}=\frac{6^3}{3!}e^{-6}=36e^{-6}.\\ P\{\xi=4\}=\frac{6^4}{4!}e^{-6}=54e^{-6}.\\ P\approx 0.715.$