MHB Normal and exponential distribution

AI Thread Summary
The discussion focuses on two statistical problems involving normal and exponential distributions. For the first problem, the mass of chocolate bars is normally distributed, and to ensure that 93% of the bars fall within a specific weight range, the calculated bound is c = 3.64 grams. The second problem involves a patient waiting in a doctor's office, where the waiting time is exponentially distributed. After waiting 5 minutes, the patient needs to wait an additional 11.5 minutes to have a 90% probability of being treated, leading to a total expected wait time of 16.5 minutes. The participants clarify the application of the memoryless property in the context of the exponential distribution.
mathmari
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Hey! :o

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)
 
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mathmari said:
2) The patient waited $5$ minutes without being called. How long does he have to wait with probability $0.9$?

Do we maybe use here the memoryless property?

It holds that $P(X>s+t|X>t) = P(X>s)$.

The patient is waiting already for $t = 5$ minutes. The probability that he will wait in total more than $s+t = s+5$ minutes is equal to $0,9$.
So, we get $$0,9=P(X>s+5 \mid X>5)=P(X>s)\Rightarrow 0.9=e^{-0.2s}\Rightarrow \ln 0.9=-0.2s\Rightarrow s\approx 0.526803$$
So, in total the patient will wait in total $s+t=0.526803+5=5.526803$ minutes? I am not really sure if this is correct, since at the first part I have calculated that within the first $11.5$ minutes the patient will be treated with probability $0.9$ and now we find that the patient has to wait in total about $5.5$ minutes with probability $0.9$ (Wondering)

Or are these two questions independent? (Wondering)
 
mathmari said:
From the Table we see that it must be $\frac{c}{2}=1,82$, so we get $c=3,64$.

Within the first $11.5$ minutes the patient will be treated with probability $0.9$.

Hey mathmari!

It looks correct to me. (Nod)

mathmari said:
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?

mathmari said:
Do we maybe use here the memoryless property?

It holds that $P(X>s+t|X>t) = P(X>s)$.

The patient is waiting already for $t = 5$ minutes. The probability that he will wait in total more than $s+t = s+5$ minutes is equal to $0,9$.
So, we get $$0,9=P(X>s+5 \mid X>5)=P(X>s)\Rightarrow 0.9=e^{-0.2s}\Rightarrow \ln 0.9=-0.2s\Rightarrow s\approx 0.526803$$
So, in total the patient will wait in total $s+t=0.526803+5=5.526803$ minutes? I am not really sure if this is correct, since at the first part I have calculated that within the first $11.5$ minutes the patient will be treated with probability $0.9$ and now we find that the patient has to wait in total about $5.5$ minutes with probability $0.9$

Or are these two questions independent?

Didn't you calculate how long he has to wait at least after those 5 minutes have past to be treated with probability 0.9? (Wondering)

It seems to me that instead we need to calculate within which total time he gets treated with probility 0.9.
At least that is how I interpret it, but I could be wrong.
If so, shouldn't we solve $t$ from $P(X < t \mid X > 5) = 0.9$? (Wondering)

Note that we can still use the memoryless property.
The time we have to wait after 5 minutes without treatment is exactly the same as if we had not waited at all. So the total time is just 5 minutes longer. (Nerd)
 
I like Serena said:
Didn't you calculate how long he has to wait at least after those 5 minutes have past to be treated with probability 0.9? (Wondering)

It seems to me that instead we need to calculate within which total time he gets treated with probility 0.9.
At least that is how I interpret it, but I could be wrong.
If so, shouldn't we solve $t$ from $P(X < t \mid X > 5) = 0.9$? (Wondering)

Note that we can still use the memoryless property.
The time we have to wait after 5 minutes without treatment is exactly the same as if we had not waited at all. So the total time is just 5 minutes longer. (Nerd)

So, is $t=11,5+5$ ? That means no matter how long he has waited before, from now we has to wait for about $11,5$ minutes?
(Wondering)
 
mathmari said:
So, is $t=11,5+5$ ? That means no matter how long he has waited before, from now we has to wait for about $11,5$ minutes?

More precisely, from now he has to wait at most 11.5 minutes until treatment with probability 0.9, giving a total time of 5 + 11.5 = 16.5 minutes.
The time he can expect to wait is different. (Nerd)
 
I like Serena said:
More precisely, from now he has to wait at most 11.5 minutes until treatment with probability 0.9, giving a total time of 5 + 11.5 = 16.5 minutes.
The time he can expect to wait is different. (Nerd)

Ah ok! Thank you so much! (Yes)
 

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