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[SOLVED] Probability that we get totally at most 30 times "head"

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Hey!! :eek:

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*}
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,686
Hey mathmari !!

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

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,686
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)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
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:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,686
Looks good to me! (Nod)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004