# Normal and exponential distribution

• MHB
• mathmari
In summary, the conversation discusses two different scenarios involving probabilities. The first scenario involves determining the bound $c$ in order for $93\%$ of chocolates to have a mass in the interval $[\mu - c, \mu + c]$. Through calculations using the normal distribution, it is found that $c=3.64$. The second scenario involves a patient waiting for treatment in a doctor's office. It is assumed that the patient's waiting time follows an exponential distribution with parameter $\lambda = 0.2$. The conversation discusses finding the time at which the patient will be treated with probability $0.9$, as well as the total time the patient can expect to wait. Using the memoryless property, it is
mathmari
Gold Member
MHB
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)

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)

## 1. What is the difference between normal and exponential distribution?

Normal distribution is a probability distribution that is symmetric around the mean, with most values falling close to the mean and fewer values farther away. Exponential distribution is a probability distribution that describes the time between events in a Poisson process, with a constant hazard rate.

## 2. How are the shapes of normal and exponential distributions different?

Normal distribution has a bell-shaped curve, while exponential distribution has a positively skewed curve that approaches zero on the left side and extends indefinitely to the right.

## 3. What is the central limit theorem and how does it relate to normal distribution?

The central limit theorem states that as the sample size increases, the sampling distribution of the sample mean will approach a normal distribution, regardless of the shape of the population distribution.

## 4. What is the use of normal distribution in real-world applications?

Normal distribution is commonly used in statistics and probability theory to model natural phenomena such as heights, weights, IQ scores, and many other variables. It is also used in quality control, finance, and other fields to analyze and make predictions based on data.

## 5. How can exponential distribution be applied in real-life situations?

Exponential distribution is commonly used to model the time between events in systems or processes that exhibit a constant failure rate, such as equipment failures, waiting times in queuing systems, and arrival times of customers or phone calls. It is also used in survival analysis to estimate the time to failure for a particular event or process.

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