02402 · Test Quiz 2

where $X$ follows a binomial with $n=200$ and $p=0.004$. Hence the correct answer is: $P(X\leq 2)=95\%$. (Most likely in reality the sampling will be done WITHOUT replacement - and hence the proper distribution would have been the Hypergeometric distribution, BUT we are only told that the total amount of lots $N$ out of which we sample the $n=200$ is a large number - many thousands - NOT what it is - which we would have needed to use the Hypergeometric distribution. And we are saved by the fact that in such cases where $n\leq N/10$, the binomial is a good approximation of the hypergeometric.

The ordered list of the $n=5$ numbers is: $80.8,80.9,82.5,83.7,95.2$. The middle number (median) is number 3 in this ordered list: $82.5$. The upper quartile is the 75% percentile, and since $n\cdot 0.75=3.75$ this number is the 4th in the ordered list: $83.7$. hence the answer is: Median = 82.5 and upper quartile=83.7 Had this been done in R, we would get: > quantile(c(82.5,83.7,80.9,95.2,80.8),type=2) 0% 25% 50% 75% 100% 80.8 80.9 82.5 83.7 95.2

The expected number of arrivals per 15 min, and the probability that in a period of 1 hour 8 guests or more arrive are? First we need to find the intensity $\lambda_{15min}$ of the poisson distribution for $X$ - we use the general expression for getting zero events in a poisson: \(P(X=0)=\frac{\lambda_{15min}^0 e^{-\lambda_{15min}}}{0!}=e^{-\lambda_{15min}}\) And since this is $0.3$ we can find \(\lambda_{15min}=-\log(0.3)=1.2\) Since $\mu=\lambda$ in a poisson, this number also expresses the expected(average) number of arrivals during 15 minutes. The second part we find by finding the intensity (average) for a 60 minutes period: \(\lambda_{60min}=4\cdot \lambda_{15min}=4.8\) And hence: \(P(X\geq 8)=1-P(X \leq 7)=1-0.887=0.113\) to be found in R as 1-ppois(7,4.8). Henc the correct answer is 2: $ \mu = 1.2$ arrivals per 15 min and $P(X_{60min} \ge 8)\approx 11\% $

The random variable X describes the number of ring binders with statistics exercises among the 3 chosen ones. The mean and variance for the random variable X is? This is a Hypergeometric distribution with $a=2$, $N=9$, $n=3$. The mean and variance of this distribution is given in Chapter 2: \(\mu=n\cdot \frac{a}{N}=3\cdot \frac{2}{9}=\frac{2}{3}\) \(\sigma^2=n\cdot \frac{a}{N}(1-\frac{a}{N})(\frac{N-n}{N-1})=3\cdot \frac{2}{9}(\frac{7}{9})(\frac{9-3}{9-1})=\frac{7}{18}\) Hence the correct answer is: $\mu\approx 0.67$ and $\sigma^2\approx 0.39$

The probability ($ {P_1} $) that all the three chosen ring binders contain reports and the probability ($ {P_2} $) to chose exactly one of each kind of ring binder are? Since there are $a=4$ that contain reports (and 5 that don’t) we use this hypergeometric distribution ($Y$) to find $P_1:$ \(P_1=P(Y=3)=\frac{\left(\begin{array}{c}4 \\ 3\end{array}\right)\left(\begin{array}{c}5 \\ 0\end{array}\right)}{\left(\begin{array}{c}9 \\ 3\end{array}\right)}=\frac{1}{21}\) For $P_2$ we use the same principle as for the hypergeometric - the basic probability approach - the denominator counts the total number of possibilities - the numerator counts the ones corresponding to the event in question. And for the latter one of each has to be chosen: \(P_2=\frac{\left(\begin{array}{c}2 \\ 1\end{array}\right)\left(\begin{array}{c}3 \\ 1\end{array}\right)\left(\begin{array}{c}4 \\ 1\end{array}\right)}{\left(\begin{array}{c}9 \\ 3\end{array}\right)}=\frac{6}{21}\)

We must first find $\lambda$, and since (using the poisson probability formula) \(0.05=P(X=0)=\frac{\lambda^0\cdot e^{-\lambda}}{0!}=e^{-\lambda}\) we get that \(\lambda=-\log(0.05)=2.9957=3.0\) And then $P(X>5)=1-P(X\leq 5)$ can be found in R as: > 1-ppois(5,3) [1] 0.08391794 And hence the answer is: $\mu=3.0$ and $P(X>5)=0.084$