# Statistics, Central limit theorem

The amount of nicotine in Dytusoon cigarettes have a mean of 0.941 g and a standard deviation of 0.313 g. Find the probability of randomly selecting 40 cigarettes with a total of 36 g of nicotine or more.

When using my TI84 calculator I enter the problem as "normalcdf(36,E99,.941,.313/(squareroot of 40))" but I get 0 as the answer. So I am wondering how I would solve this using my calculator( what part of my formula is wrong or is there a step I am missing).

• Welcome to MSE. You've posted a textbook exercise, but unfortunately, you seem to have neglected to ask your question. Please edit accordingly. – T. Bongers Jul 19 '16 at 3:32
• Find the probability of randomly selecting 40 cigarettes with a total of 36 g of nicotine or more. – rickmath Jul 19 '16 at 3:42
• It seems you've misunderstood my comment. Copy-pasting the exercise again is not asking the question that you have, unless you just want someone to do the problem for you (which I'm sure you don't). What concept are you having trouble applying here? – T. Bongers Jul 19 '16 at 3:44
• When using my TI84 calculator I enter the problem as "normalcdf(36,E99,.941,.313/(squareroot of 40))" but I get 0 as the answer. So I am wondering how I would solve this using my calculator( what part of my formula is wrong or is there a step I am missing). – rickmath Jul 19 '16 at 3:49

My guess is that you aren't calculating the probability associated with the correct distribution, so the answer you are getting is too small, so you are getting an underflow error and thus your calculator reports 0. Let's rehash the steps to solving the problem here

EDIT:

The first entry probably should not be 36. I can't tell you exactly what I think it might be since I don't know how the "normalcdf" function on your calculator works or what its parameters are; there are multiple ways to interpret it.

My guess is that the "normalcdf" function is using the distribution of $\mathcal{N}(0,1)$ to calculate the probabilities.

However, what you want to calculate is the distribution of $\mathcal{N}(40*(0.941),40*(0.313)^2)$ where the first number in parentheses is the mean, and the second is the variance (square of the standard deviation). We can transform the probabilities of the one distribution to another as follows (I believe; I am doing this all in my head and not writing it down so I am probably making mistakes).

What we want to calculate is $\mathbb{P}(Z \ge 36)$ where $Z = \sum_{i=1 }^{40} X_i$, and $Z \sim \mathcal{N}(40*(0.941),40*(0.313)^2)$ but your calculator only lets you calculate the distribution of $\mathcal{N}(0,1)$. So we normalize $Z$ to get a random variable $$Y= \frac{Z-(40*0.941)}{\sqrt{40}*(0.313)}$$ which is distributed as $\mathcal{N}(0,1)$.

Now $$\mathbb{P}(Z \ge 36)=\mathbb{P}\left(Y \ge \frac{36-40*0.941}{\sqrt{40}*0.313}\right)$$

Let $X$ be the amount of nicotine in the $i$th cigarette, $i=1, \dots, 40$.

Assume that the amount of nicotine in each cigarette is independent, i.e. all of the $X_i$ are independent.

Using the Central Limit Theorem, approximate the $X_i$ by normal variables with mean $0.941$ and and variance $0.313$.

Thus what you want to calculate is the probability distribution of $\sum_{i=1}^{40} X_i$, which is the sum of independently distributed normal random variables, and therefore normally distributed itself.

Make sure you are calculating the mean and variance of $\sum_{i=1}^{40} X_i$ correctly; they are not the same as the mean and variance of the $X_i$, which looks like what you calculated above. See the formulas on Wikipedia for two independently distributed normal random variables, https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables, and then generalize.

• Thank you for the response, I am a little lost from what you said. We dont have a book for the class and my teacher only shows us how to enter information into our calculator and not how to work it out on paper so I am lost when it comes to the terms :((. would you be able to explain it in steps of using a calculator. – rickmath Jul 19 '16 at 4:04
• That looks like it, I was trying to figure out how to reduce 36 earlier but didn't think of dividing it by 40. Thank you William! – rickmath Jul 19 '16 at 4:20
• OK take the suggestion with a grain of salt though because I don't know how your calculator's function works, so I could still be telling you to pass the wrong parameter - I really am not quite sure. – Chill2Macht Jul 19 '16 at 4:27

Hint: The sum of normally distributed and independent random variables are normally distributed as well.

If $X_i\sim \mathcal N(\mu, \sigma ^2)$

then $\sum_{i=1}^n X_i\sim \mathcal N(n\cdot \mu, n\cdot \sigma ^2)$

Therefore $P(\sum_{i=1}^{40} X_i\geq 36)=1-P(\sum_{i=1}^{40} X_i <36)$

$=1-\Phi\left(\frac{36-40\cdot 0.941}{\sqrt{40} \cdot 0.313}\right)=1-\Phi(-0.82846)$

where $\Phi(z)$ is the cdf of the standard normal distribution. The value for this function can be looked up in this or other tables.

If you use a calculator with the function $\texttt{normalcdf}(X; \mu; \sigma)$, then you get the result of $P(\sum_{i=1}^{40} X_i\geq 36)$ by typing in $\texttt{normalcdf}(36;40\cdot 0.941; \sqrt{40}\cdot 0.313)$ or

$\texttt{normalcdf}(36; 37.64; 1.9796)$

Then you take the result and subtract it from $1$.