# Central Limit Theorem prove an expectation could be wrong

I really need your help. Here's the problem:

A rookie is brought to a baseball club on the assumption that he will have a .300 batting average. (Batting average is the ratio of the number of hits to the number of times at bat.) In the first year, he comes to bat 300 times and his batting average is .267. Assume that his at bats can be considered Bernoulli trials with probability .3 for successs. Could such a low average be considered just bad luck or should he be sent back to the minor leagues?

I know that, as his batting average was .267 in 300 times, he only hit 80 times.

I know I have to use the Central Limit Theorem, but how? First I thought I had to calculate, for example, the probability that he had hit between 75 to 85 times. Obviously knowing that this can be modeled by a Binomial(300,0.3) and aproximating this by a Normal (because n=300). So I could use the CLT that way.

But I think that what I really need to do is to give an interval, where I can assure, with high probability (90% for example), the real expectation is in that interval.
So, how I bound the real expectation using the CLT?

(Sorry if I wrote something wrong, English is not my first language)

• are you sure you understand problem correctly? For me it states that 300 "at bats" number is a random one as well as number of hits. May 12, 2015 at 1:04
• @aandreev , I was confused by that sentence too. I think that sentence "his at bats can be considered Bernoulli" is intended to mean that every time he is at bat, his success is a Bernoulli $0/1$ trial. We are given the number of at bats is (deterministically) 300. May 12, 2015 at 3:36

A batting average of $.267$ is pretty good in the major leagues.

But for your probability problem, I think you want to do something like this: Let $X_i$ be the random Bernoulli outcome for batting attempt $i$, for $i \in \{1, \ldots, 300\}$. Assume $\{X_1, \ldots, X_{300}\}$ are mutually independent Bernoulli with $Pr[X_i=1]=0.3$. So you want to compute $Pr[\sum_{i=1}^{300} X_i \leq 80]$. You can translate this into a CLT problem by:

\begin{align} Pr\left[\sum_{i=1}^{300} X_i \leq 80\right] &= Pr\left[\frac{1}{\sqrt{300 Var(X_1)}}\sum_{i=1}^{300} (X_i-.3)\leq \frac{1}{\sqrt{300 Var(X_1)}}(80-90)\right]\\ &\approx Pr\left[G \leq \frac{-10}{\sqrt{300 Var(X_1)}}\right] \end{align} where $G$ is a standard Normal Gaussian with zero mean and unit variance, and this assumes that 300 samples is "large enough" for the approximation to be good. So now use a "lookup table approach." By symmetry of the Gaussian, this is also equal to the $Q(\cdot)$ function evaluated at $10/\sqrt{300 Var(X_1)}$, where $Q(x) = \int_{x}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-t^2/2}dt$.

"Lookup tables" are so outdated, I don't know why they are still taught when everyone has a computer at their disposal. Also, I don't know why computers and calculators do not have a standard $Q(\cdot)$ function built in, they have the much less useful $erf(\cdot)$ function, and I always have to try to remember how that one is defined and how to derive one from the other.

Okay, I looked it up. We have the following definitions for $Q(\cdot)$ and $erf(\cdot)$:

\begin{align} Q(x) &= \frac{1}{\sqrt{2\pi}} \int_x^{\infty} e^{-t^2/2}dt\\ erf(x) &= \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}dt \end{align}

and so: $$Q(x) = \frac{1}{2}-\frac{1}{2}erf\left(\frac{x}{\sqrt{2}}\right)$$

• Thanks for answering. I had done that the result is 0.0869(aprox.) May 12, 2015 at 16:11
• But my question is why do we have to calculate Pr[∑300i=1Xi≤80]?, why don't we calculate Pr[75≤∑300i=1Xi≤85] (for example) ? That's why I want to "bound" the real expectation with a high probability. And with that, I could answer if 0.3 could be the real expectation or not. May 12, 2015 at 16:18
• I get a different answer of $0.1039$. If you want to bound the probability we are in the interval $[75, 85]$ you can do that, but that seems arbitrary. The above gives you an approximation of the probability of being less than or equal to 80, assuming the .3 probability is correct. I guess if you get a small number you can decide whether or not it is more likely for that to happen, or for the person to lie about his batting average. May 12, 2015 at 18:03
• You could be more formal and assume he has 2 possible batting averages, $0.3$ and $0.15$, and the probability of being $0.15$ is the probability of lying, which is $1/2$. Define $\theta = -10/\sqrt{300Var(X_1)}$. Then $Pr[lie | G=\theta] = f_G(\theta | lie)(1/2)/[f_G(\theta|lie)(1/2) + f_G(\theta|true)(1/2)]$. May 12, 2015 at 18:19
• Note also that if you want to give an interval for which it is likely to be in that interval, the highest probability comes from the largest interval that is relevant, which is $(80, \infty]$ for this problem. May 12, 2015 at 18:29