# Finding mean and total probabilities

A population of widgets produced by a machine has weights with mean of 200mg and standard deviation of 10mg. If a random sample of 36 was taken,

a) What is the probability that the mean weight is at least 196mg?

b) What is the probability that the total weight is at most 3100mg?

I think I'm on the right trick with part a), but my approach to part b) must contain some kind of mistake...

a) Standardizing...

$E(M_{36}) = 200$

$var(M_{36}) = \frac{10^2}{36} = \frac{25}{9}$

$P(M_{36} \geq 196) = P(\frac{M_{36} - 200}{\sqrt\frac{25}{9}} \geq \frac{196 - 200}{\sqrt\frac{25}{9}}) = P(Z \geq \frac{196 - 200}{\frac{5}{3}}) = P(Z \geq \frac{196 - 200}{\frac{5}{3}}) = P(Z \geq -2.4) = 1 - P(Z < 2.4) = 1 - .9918 = .0082$

b) Standardizing...

$E(S_{36}) = 36 * 200 = 7200$

$var(S_{36}) = 36 * 10^2 = 3600$

$P(S_{36} \leq 3100) = P(\frac{S_{36} - 7200}{\sqrt{3600}} \leq \frac{3100 - 7200}{\sqrt{3600}}) = P(Z \leq \frac{3100 - 7200}{60}) = P(Z \leq \frac{-205}{3})$

This is where I get stuck.

• So far so good. The number in part b) is far away from mean, in terms of the standard score you computed, so you may answer something like approximately zero. – BGM Oct 10 '17 at 11:22
• Do we know that the distribution of weights for the population of widgets is normal, or do we just know that the distribution has the given mean and standard deviation? – MMASRP63 Oct 12 '17 at 3:14
• Your calculations are fine. It is simply extremely unlikely that the total weight does not exceed 3100 mg. – H. H. Rugh Oct 12 '17 at 17:15

This, definitely, is not a complete answer, but I believe it might be helpful. The key point, I would say, is to understand how your samples are drawn, i.e., to define the joint distribution over 36 random variables
\begin{align} p(x_{1},\cdots,x_{36}), \label{eq:joint} \end{align} then your problems can be formalized through expectations and solved using Chebychev's inequality. That is,

(a) \begin{align} E[g(x_{1},\cdots,x_{36})] = \int g(x_{1},\cdots,x_{36})p(x_{1},\cdots,x_{36}) \end{align} where \begin{align} g(x_{1},\cdots,x_{36}) = \frac{1}{36}\sum^{36}_{i=1} x_{i} \end{align} (b) is define as \begin{align} E[f(x_{1},\cdots,x_{36})] = \int f(x_{1},\cdots,x_{36})p(x_{1},\cdots,x_{36}) \end{align} where \begin{align} f(x_{1},\cdots,x_{36}) =\sum x_{i} \end{align}

Hint on solving both equations:

The probablity of the samples can be factored as $p(x_{1},\cdots,x_{36}) = p(x_{1}\mid x_{2},\cdots,x_{36}),\cdots,p(x_{35}|x_{36})$. Now, this amounts to knowing how the samples are drawn. e.g., are the samples drwan independnetly of each other. Using such kind of prior knowldege about your samples, you can try to compute the expectations given in (a) and (b) and use the Chebychev’s inequality to solve for you specific problem. Fo instance,

\begin{align} p(f(x_{1},\cdots,x_{36}) \geq 3100) \leq \frac{E[f(x_{1},\cdots,x_{36})]}{3100} \end{align}

which you can use to compute the probability of it being $\leq$ 3100 directly.

Good luck.

Your calculations in both parts are correct, assuming that the population is Normal; otherwise, these are Normal approximations relying on the Central Limit Theorem and the fact that the term "random sample" ordinarily refers to observations that are independent and identically distributed (i.i.d.) random variables.

Note that $P(Z \leq \frac{-205}{3})<<0.0001$: as you may recall, more than $99.99$% of any Normal distribution falls within $4$ standard deviations of the mean, whereas here we have $\frac{205}{3}= 68\frac{1}{3}$ standard deviations!

Also, note that $S_n = n\,M_n,$ so when you standardize $S_n$ you get the same random variable $Z$ as when you standardize $X_n$:

$$\frac{S_n - E[S_n]}{\sqrt{V[S_n]}}=\frac{n\,M_n - E[n\,M_n]}{\sqrt{V[n\,M_n]}}=\frac{n\,M_n - n\,E[M_n]}{\sqrt{n^2\,V[M_n]}}=\frac{M_n - E[M_n]}{\sqrt{V[M_n]}}=Z.$$

In other words, both of your probabilities pertain to the distribution of $Z$, which it appears you are expected to assume is at least approximately Standard Normal.

You are right in the answer to part (b). However, I think the answer to part (a) is incorrect.

Part a:

Part (a) asks the probability that the sample mean is in the range [true mean - 4mg, true mean + $\infty$mg] This amswer should be more than 50% since it fully covers the higher side of the true mean (which is 50%) plus some part on the lower side.

$P(Z≥−2.4)=1−P(Z<-2.4)$ you missed the minus sign above.

We should have $P(Z≥−2.4)=P(Z<2.4)$ (since the distribution is symmetric). Finally,we have $P(Z<2.4) =.9918$

So, answer to part a is $.9918$.

Part b Your derivation is right. The last step is to evaluate the probability $P(Z < -K)$ where $K$ is a large positive number.

Now, $P(Z<-K)=P(Z>K)$. This is very close to zero, so don't try plugging this in directly into some function call. A common approximation is the following: $P(Z>K)\leq e^{-K^2/2}/(K\sqrt{2\pi})$ Your class notes might give you a different approximation - you can use that.
For part b, we need to evaluate the above for $K=205/3=68.33$. $e^{-K^2/2}=e^{-2334}=10^{-1014.}$

Answer to part b is $10^{-1014}/171.2$ For all practical purposes, you can treat this as zero.