Chapter 1, Problem 1.10P

(a)

To determine

The probability of getting each of the 10 digits by picking one from the first 25 digits in the decimal expansion of $\pi$.

Answer to Problem 1.10P

The probability of getting each of the 10 digits are $0,\frac{2}{25},\frac{3}{25},\frac{5}{25},\frac{3}{25},\frac{3}{25},\frac{3}{25}\frac{1}{25},\frac{2}{25},\frac{3}{25}$ respectively.

Explanation of Solution

Write the expansion of $\pi$

  $\pi =3.141592653589793238462643$

Write the expression for probability

    $P\left(j\right)=\frac{N\left(j\right)}{N}$        (I)

Here, $P\left(j\right)$ is the probability of getting the digit $j$, $N\left(j\right)$ is the number of times the digit $j$ occurs and $N$ is the total number of digits.

Conclusion:

Substitute $0$ for $N\left(0\right)$ and $25$ for $N$ to find $P\left(0\right)$

  $\begin{array}{c}P\left(0\right)=\frac{0}{25}\\ =0\end{array}$

Substitute $2$ for $N\left(1\right)$ and $25$ for $N$ to find $P\left(1\right)$

  $P\left(1\right)=\frac{2}{25}$

Substitute $3$ for $N\left(2\right)$ and $25$ for $N$ to find $P\left(2\right)$

  $P\left(2\right)=\frac{3}{25}$

Substitute $5$ for $N\left(3\right)$ and $25$ for $N$ to find $P\left(3\right)$

  $P\left(3\right)=\frac{5}{25}$

Substitute $3$ for $N\left(4\right)$ and $25$ for $N$ to find $P\left(4\right)$

  $P\left(4\right)=\frac{3}{25}$

Substitute $3$ for $N\left(5\right)$ and $25$ for $N$ to find $P\left(5\right)$

  $P\left(5\right)=\frac{3}{25}$

Substitute $3$ for $N\left(6\right)$ and $25$ for $N$ to find $P\left(6\right)$

  $P\left(3\right)=\frac{3}{25}$

Substitute $1$ for $N\left(7\right)$ and $25$ for $N$ to find $P\left(7\right)$

  $P\left(7\right)=\frac{1}{25}$

Substitute $2$ for $N\left(8\right)$ and $25$ for $N$ to find $P\left(8\right)$

  $P\left(9\right)=\frac{2}{25}$

Substitute $3$ for $N\left(9\right)$ and $25$ for $N$ to find $P\left(9\right)$

  $P\left(9\right)=\frac{3}{25}$

Therefore, the probability of getting each of the 10 digits are $0,\frac{2}{25},\frac{3}{25},\frac{5}{25},\frac{3}{25},\frac{3}{25},\frac{3}{25}\frac{1}{25},\frac{2}{25},\frac{3}{25}$ respectively.

(b)

To determine

The most probable digit, the median digit and the average value.

Answer to Problem 1.10P

The most probable digit is $3$, the median digit is $4$  and the average value is $4.72$.

Explanation of Solution

The most probable digit is the digit for which $P\left(j\right)$ is maximum. From part (a), $P\left(j\right)$ is maximum for $j=3$.

The median is the value of $j$ such that $P\left(X\ge j\right)=P\left(X\le j\right)$.

Write the expression for the average value.

  $\langle j\rangle =\frac{{\displaystyle \sum jN\left(j\right)}}{N}$        (II)

Here, $\langle j\rangle$ is the expectation value of $j$.

Conclusion:

From part (b), write the expression for $P\left(X\le 4\right)$.

  $\begin{array}{c}P\left(X\ge 4\right)=\frac{0}{25}+\frac{2}{25}+\frac{3}{25}+\frac{5}{25}+\frac{3}{25}\\ =\frac{13}{25}\end{array}$

From part (b), write the expression for $P\left(X\ge 4\right)$.

  $\begin{array}{c}P\left(X\le 4\right)=\frac{3}{25}+\frac{3}{25}+\frac{3}{25}+\frac{1}{25}+\frac{2}{25}+\frac{3}{25}\\ =\frac{15}{25}\end{array}$

Expand the summation in equation (II) and substitute the corresponding values.

  $\begin{array}{c}\langle j\rangle =\frac{\left[0\left(0\right)+1\left(2\right)+2\left(3\right)+3\left(5\right)+4\left(3\right)+5\left(3\right)+6\left(3\right)+7\left(1\right)+8\left(2\right)+9\left(3\right)\right]}{25}\\ =4.72\end{array}$

Therefore, the most probable digit is $3$, the median digit is $4$ and the average value is $4.72$.

(c)

To determine

The standard deviation of the given distribution.

Answer to Problem 1.10P

The standard deviation of the given distribution is $2.474$.

Explanation of Solution

Write the expression for the expectation value of $j^2$

  $\langle j^2\rangle =\frac{{\displaystyle \sum j^{2}N\left(j\right)}}{N}$        (III)

Here, $\langle j^2\rangle$ is the expectation value of $j$.

Write the expression for standard deviation

  $\sigma =\sqrt{\langle j^2\rangle -{\langle j\rangle }^2}$        (IV)

Here, $\sigma$ is the standard deviation.

Conclusion:

Expand the summation in equation (II) and substitute the corresponding values

  $\begin{array}{c}\langle j^2\rangle =\frac{\left[0^2\left(0\right)+1^2\left(2\right)+2^2\left(3\right)+3^2\left(5\right)+4^2\left(3\right)+5^2\left(3\right)+6^2\left(3\right)+7^2\left(1\right)+8^2\left(2\right)+9^2\left(3\right)\right]}{25}\\ =28.4\end{array}$

Substitute $28.4$ for $\langle j^2\rangle$ and $4.72$ for $\langle j\rangle$ in equation (III) to find $\sigma$.

  $\begin{array}{c}\sigma =\sqrt{\left(28.4\right)-{\left(4.72\right)}^2}\\ =2.474\end{array}$

Therefore, the standard deviation of the given distribution is $2.474$.