MHB Probability that we get totally at most 30 times "head"

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The discussion revolves around the probability of obtaining heads when tossing a fair coin multiple times. Each toss is modeled as a Bernoulli random variable, and the average proportion of heads is analyzed using the central limit theorem. To achieve a 90% probability that the proportion of heads falls between 0.45 and 0.55, at least 273 tosses are required. Additionally, when tossing the coin 100 times, the probability of getting at most 30 heads is calculated, resulting in a value close to zero. The calculations and assumptions made throughout the discussion are confirmed to be correct.
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Hey! :o

A fair coin is tossed $ n $ times; $ X_i = 1 $ denotes the event that "head" appears in the $ i $-th toss.

a) How are the single toss $X_i$, $=1, \ldots , n$, distributed?

b) How many toss are needed so that the proportion of "head" $\overline{X}_n$ is in the interval $0, 45 < \overline{X}_n < 0, 55$ with probability $90\%$ ?

c) Given that the coin is tossed $100$ times, determine the probability that we get totally at most $30$ times "head".

I have done the following:

a) Each $X_i$ is Bernoulli distributed, since we either have success or not.

b) Let $\overline{X} = \frac{1}{n}(X_1 + X_2+ \ldots +X_n)$. According to the zentral limit theorem $\overline{X}$ approximates the normal distribution for large $n$ with parameters \begin{equation*}E(\overline{X})=\frac{1}{2}=0.5 \ \text{ and } \ V(\overline{X})=\frac{\sigma_X^2}{n}=\frac{\left (\frac{1}{4}\right )^2}{n}=\frac{1}{16n}\Rightarrow \sigma_{\overline{X}}=\frac{1}{4\sqrt{n}}\end{equation*}

Therefore, so that it hods $0, 45 < \overline{X}_n < 0, 55$ with probability $90\%$, we have the following:
\begin{align*}P\left (0,45<\overline{X}_n<0,55\right )\geq 90\% &\Rightarrow \Phi \left (\frac{0.55-0.5}{\frac{1}{4\sqrt{n}}}\right )-\Phi \left (\frac{0.45-0.5}{\frac{1}{4\sqrt{n}}}\right )\geq 0.9 \\ & \Rightarrow \Phi \left (0.2\sqrt{2}\right )-\Phi \left (-0.2\sqrt{n}\right )\geq 0.9\\ & \Rightarrow 2\Phi \left (0.2\sqrt{2}\right )\geq 1.9 \\ & \Rightarrow \Phi \left (0.2\sqrt{2}\right )\geq 0.95 \\ & \Rightarrow 0.2\sqrt{n}\geq 1.65 \\ & \Rightarrow n\geq \frac{1089}{16}=8.25 \end{align*}

So, we need at least 9 toss.
Is everything correct s far? At c) do we define the random variable $Z:=n\cdot \overline{X}_n$ with $n=100$ and calculate the probability $P(Z\leq 30)$ ?

If yes, I must have somewhere a mistake, because I get the following:

We have that \begin{equation*}E(Z)=E(100\overline{X}_{100})=100\cdot E(\overline{X}_{100})=100\cdot 0.5=50\end{equation*} nd \begin{equation*}V(Z)=Z(100\overline{X}_{100})=100^2\cdot V(\overline{X}_{100})=10000\cdot \frac{1}{16\cdot 100}=\frac{25}{4}=6.25\end{equation*}

So $Z\sim N(50, 6.25)$.

The probability that we are looking for is then equal to \begin{equation*}P(Z\leq 30)=\Phi \left (\frac{30-50}{\sqrt{6.25}}\right )=\Phi \left (\frac{-20}{2.5}\right )=\Phi \left (-8\right )=1-\Phi (8)\end{equation*}
 
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Hey mathmari!

That looks all correct to me.
How come that you think there is a mistake? (Wondering)
 
I like Serena said:
That looks all correct to me.
How come that you think there is a mistake? (Wondering)

I thought so because we have $\Phi (8)$ so a very big argument, that doesn't exist in a table of normal distribution. At such a table the largest number converge to $1$. Does this mean that $\Phi (8)\approx 1$ and so $P(Z\leq 30)\approx 1-1=0$ ? (Wondering)
 
mathmari said:
b) Let $\overline{X} = \frac{1}{n}(X_1 + X_2+ \ldots +X_n)$. According to the zentral limit theorem $\overline{X}$ approximates the normal distribution for large $n$ with parameters \begin{equation*}E(\overline{X})=\frac{1}{2}=0.5 \ \text{ and } \ V(\overline{X})=\frac{\sigma_X^2}{n}=\frac{\left (\frac{1}{4}\right )^2}{n}=\frac{1}{16n}\Rightarrow \sigma_{\overline{X}}=\frac{1}{4\sqrt{n}}\end{equation*}

Shouldn't we have $\sigma_X^2=\frac 14$ instead of $\sigma_X^2 =(\frac 14)^2$? (Wondering)
 
I like Serena said:
Shouldn't we have $\sigma_X^2=\frac 14$ instead of $\sigma_X^2 =(\frac 14)^2$? (Wondering)

Oh yes. (Blush)

So, we have the following:
\begin{equation*}E(\overline{X})=\frac{1}{2}=0.5 \ \text{ and } \ V(\overline{X})=\frac{\sigma_X^2}{n}=\frac{\frac{1}{4}}{n}=\frac{1}{4n}\Rightarrow \sigma_{\overline{X}}=\frac{1}{2\sqrt{n}}\end{equation*}

So, we get that
\begin{align*}P\left (0,45<\overline{X}_n<0,55\right )\geq 90\% &\Rightarrow \Phi \left (\frac{0.55-0.5}{\frac{1}{2\sqrt{n}}}\right )-\Phi \left (\frac{0.45-0.5}{\frac{1}{2\sqrt{n}}}\right )\geq 0.9 \\ & \Rightarrow \Phi \left (0.1\sqrt{n}\right )-\Phi \left (-0.1\sqrt{n}\right )\geq 0.9 \\ & \Rightarrow \Phi \left (0.1\sqrt{n}\right )-\left [1-\Phi \left (0.1\sqrt{n}\right )\right ]\geq 0.9 \\ & \Rightarrow \Phi \left (0.1\sqrt{n}\right )-1+\Phi \left (0.1\sqrt{n}\right )\geq 0.9 \\ & \Rightarrow 2\Phi \left (0.1\sqrt{n}\right )\geq 1.9 \\ & \Rightarrow \Phi \left (0.1\sqrt{n}\right )\geq 0.95 \\ & \Rightarrow 0.1\sqrt{n}\geq 1.65 \\ & \Rightarrow n\geq 272.25\end{align*}

So, at least $273$ toss have to be done.
Then, we have that \begin{equation*}E(Z)=E(100\overline{X}_{100})=100\cdot E(\overline{X}_{100})=100\cdot 0.5=50\end{equation*} and \begin{equation*}V(Z)=Z(100\overline{X}_{100})=100^2\cdot V(\overline{X}_{100})=10000\cdot \frac{1}{4\cdot 100}=25\end{equation*}

Therefore, $Z\sim N(50, 25)$.

So, the probability is equal to \begin{equation*}P(Z\leq 30)=\Phi \left (\frac{30-50}{\sqrt{25}}\right )=\Phi \left (\frac{-20}{5}\right )=\Phi \left (-4\right )=1-\Phi (4)\approx 1-1=0\end{equation*}
Is everything correct now? (Wondering)
 
Last edited by a moderator:
Looks good to me! (Nod)
 
I like Serena said:
Looks good to me! (Nod)

Great! Thank you very much! (Yes)
 

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