- #1
mathmari
Gold Member
MHB
- 4,984
- 7
Hey! 
I am looking at the following:
1) A machine produces $100$ gram chocolate. Due to random influences, not all bars are equally heavy. From a long series of observations it is known that the mass X of a chocolate is distributed normally with parameters $\mu = 100$g and $\sigma = 2.0$g.
How do we have to choose the bound $c$ so that $93\%$ of chocolates have a mass in the intervall $[\mu - c, \mu + c]$ ?
I have done the following:
\begin{align*}P(\mu-c\leq X\leq \mu+c)=0,93&\Rightarrow \Phi\left (\frac{\mu+c-\mu}{\sigma}\right ) -\Phi\left (\frac{\mu-c-\mu}{\sigma}\right )=0,93 \\ & \Rightarrow \Phi\left (\frac{c}{\sigma}\right ) -\Phi\left (\frac{-c}{\sigma}\right )=0,93 \\ & \Rightarrow \Phi\left (\frac{c}{\sigma}\right ) -\left (1-\Phi\left (\frac{c}{\sigma}\right )\right )=0,93 \\ & \Rightarrow \Phi\left (\frac{c}{\sigma}\right ) -1+\Phi\left (\frac{c}{\sigma}\right )=0,93 \\ & \Rightarrow 2\cdot \Phi\left (\frac{c}{\sigma}\right ) =1,93 \\ & \Rightarrow 2\cdot \Phi\left (\frac{c}{2}\right ) =1,93 \\ & \Rightarrow \Phi\left (\frac{c}{2}\right ) =0,965 \end{align*}
From the Table we see that it must be $\frac{c}{2}=1,82$, so we get $c=3,64$.2) A patient is sitting in the waiting room of a doctor's office. We assume that its waiting time in minutes is exponentially distributed with parameter $\lambda = 0.2$. Within what time will the patient be treated with probability $0.9$?
The patient waited $5$ minutes without being called. How long does he have to wait with probability $0.9$?
I have done the following:
$$P(X\leq x)=1-e^{-\lambda x}\Rightarrow 0.9=1-e^{-0.2x} \Rightarrow e^{-0.2x} =0.1 \Rightarrow \ln e^{-0.2x}=\ln 0.1 \\ \Rightarrow -0.2x\approx -2.30259 \Rightarrow x= 11.513$$
Within the first $11.5$ minutes the patient will be treated with probability $0.9$.
Is at the second question "The patient waited $5$ minutes without being called. How long does he have to wait with probability $0.9$?" the answer "He has to wait for 11,5-5 minutes."? Or do we have to calculate something else here?
Is everything correct? Could I improve something? (Wondering)
I am looking at the following:
1) A machine produces $100$ gram chocolate. Due to random influences, not all bars are equally heavy. From a long series of observations it is known that the mass X of a chocolate is distributed normally with parameters $\mu = 100$g and $\sigma = 2.0$g.
How do we have to choose the bound $c$ so that $93\%$ of chocolates have a mass in the intervall $[\mu - c, \mu + c]$ ?
I have done the following:
\begin{align*}P(\mu-c\leq X\leq \mu+c)=0,93&\Rightarrow \Phi\left (\frac{\mu+c-\mu}{\sigma}\right ) -\Phi\left (\frac{\mu-c-\mu}{\sigma}\right )=0,93 \\ & \Rightarrow \Phi\left (\frac{c}{\sigma}\right ) -\Phi\left (\frac{-c}{\sigma}\right )=0,93 \\ & \Rightarrow \Phi\left (\frac{c}{\sigma}\right ) -\left (1-\Phi\left (\frac{c}{\sigma}\right )\right )=0,93 \\ & \Rightarrow \Phi\left (\frac{c}{\sigma}\right ) -1+\Phi\left (\frac{c}{\sigma}\right )=0,93 \\ & \Rightarrow 2\cdot \Phi\left (\frac{c}{\sigma}\right ) =1,93 \\ & \Rightarrow 2\cdot \Phi\left (\frac{c}{2}\right ) =1,93 \\ & \Rightarrow \Phi\left (\frac{c}{2}\right ) =0,965 \end{align*}
From the Table we see that it must be $\frac{c}{2}=1,82$, so we get $c=3,64$.2) A patient is sitting in the waiting room of a doctor's office. We assume that its waiting time in minutes is exponentially distributed with parameter $\lambda = 0.2$. Within what time will the patient be treated with probability $0.9$?
The patient waited $5$ minutes without being called. How long does he have to wait with probability $0.9$?
I have done the following:
$$P(X\leq x)=1-e^{-\lambda x}\Rightarrow 0.9=1-e^{-0.2x} \Rightarrow e^{-0.2x} =0.1 \Rightarrow \ln e^{-0.2x}=\ln 0.1 \\ \Rightarrow -0.2x\approx -2.30259 \Rightarrow x= 11.513$$
Within the first $11.5$ minutes the patient will be treated with probability $0.9$.
Is at the second question "The patient waited $5$ minutes without being called. How long does he have to wait with probability $0.9$?" the answer "He has to wait for 11,5-5 minutes."? Or do we have to calculate something else here?
Is everything correct? Could I improve something? (Wondering)