Central Limit Theorem to approx Binomial Distribution, cannot see how they got this answer

binomial distributioncentral limit theoremnormal distributionprobability

I'm doing question b, using the Central Limit Theorem to approximate a Binomial Distribution to calculate it's sample mean:

There are 20 children in a class. Each flips a biased coin 15 times. The probability of getting a head is 0.25.

a Write down the expected number of heads that each child would get.

b Find an estimate of the probability that the mean number of heads is at most 4.

So calculate $P(X \leq 4)$ given $N_s = 20$, $n = 15$, and $p = 0.25$.

So:

mean $= np = 0.25 \times 15 = 3.75$

variance $= np(1-p) = 3.75 \times 0.75 = 2.8125$

So use CLT with $X \sim N(3.75, 2.8125/20)$

On my calculator I use the Cumulative Normal, with lower limit $-\infty$, upper limit $4$, $\sigma=\sqrt{2.8125/20}$, $\mu=3.75$

I get $0.748$, but the textbook answer is $0.0748$:

Am I right and this is a typo?

Best Answer

OK, so you have indeed have $N=20,n=15,p=0.25$, so then $$ \begin{split} \mu &= np &= 15*0.25 &= 3.75 \\ \sigma^2 &= np(1-p) &= 15*0.25*0.75 &= 2.8125 \end{split} $$ and the standard deviation to use in the Normal distribution approximation would be $$ s = \sqrt{\sigma^2/N} = \sqrt{2.8125/20} = 0.375. $$

You can use Wolfram Alpha Normal CDF calculator to see that $\mathbb{P}[X < 4] = 0.7475$.

Related Solutions

[Math] Using central limit theorem to estimate probability

Define $X_i$ as the number of 1's on the i's toss. $X_i=1$ if the $i^{th}$ toss is 1, and $X_i=0$ if the $i^{th}$ toss is not 1.

By central limit theorem, we can assume $X_i's$ have normal distribution with $E(X)=9000\times\frac{1}{3}=3000$.

$Var(X)=9000\times(\frac{1}{3}-(\frac{1}{3})^2)=2000$

$\sigma=\sqrt{Var(X)}=\sqrt{2000}$

Let $T=\frac{\bar{X}-\mu}{\sigma}$, and $T$~$N(0,1)$.

So we want to calculate:

$P(2970<X<3040)=P(\frac{2970-3000}{\sqrt{2000}}<T<\frac{3040-3000}{\sqrt{2000}})$

Using R studio, with command:

pnorm((3040-3000)/sqrt(2000))-pnorm((2970-3000)/sqrt(2000))

We get $P(2970<X<3040)=P(\frac{2970-3000}{\sqrt{2000}}<T<\frac{3040-3000}{\sqrt{2000}})=0.5632858$

[Math] Central limit theorem – Coin toss

You can define $X_i$ as you suggest though not all Euros have a head (is it the map side they all have? or the other side which sometimes has a head?) Let's define $X_i$ as an indicator of the map side, so you either want at least $110$ or no more than $90$ map sides showing

You then want $\mathbb P \left(\sum X_i \ge 110\right)+\mathbb P \left(\sum X_i \le 90\right)$

Assuming independence, you can then use the binomial distribution for an exact calculation, or a Gaussian approximation with a cuttoffs at $90.5$ and $109.5$ (or using symmetry, double no more than $90.5$). The probability would be almost $18\%$

For example in R:

> pbinom(90, size=200, prob=0.5) + 1 - pbinom(109, size=200, prob=0.5)
[1] 0.178964

> 2 * pnorm(90.5, mean=200*0.5, sd=sqrt(200*0.5*0.5)) 
[1] 0.1791092

Best Answer

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[Math] Using central limit theorem to estimate probability

[Math] Central limit theorem – Coin toss

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