Digit Sums

Normally, we shouldn't conflate expected value with median (the most likely outcome), but in this case, they are the same.

The possible choices for the 1st digit are $1$ through $9,$ and so the expected value of the first digit is $\frac{1+9}{2}=5.$ The possible choices for all the other digits are $0$ through $9,$ and so the expected value of each other digit is $\frac{0+9}{2}=4.5.$ Then the expected value for the 7-digit number is $5+4.5\times 6=\boxed{32}.$

Generalized to $n$ digits, the expected value of the digit sum is $5+4.5n.$

Let $S$ represent the sum of the digits of a random 7-digit positive number. If $X_n$ is the value of the nth digit from left to right, then $S = X_1 + X_2 + X_3 + X_4 + X_5 + X_6 + X_7$ and by linearity of expectation $E[S] = E[X_1] + E[X_2] + E[X_3] + E[X_4] + E[X_5] + E[X_6] + E[X_7]$ . Since the distribution of $X_n$ is the same for $n \geq 2$ , $E[S] = E[X_1] + 6E[X_2]$ . The range of values for $X_1$ is $[1, 9]$ with each digit being equally likely to be chosen, so $E[X_1] = \frac{1}{9}\frac{9\left(10\right)}{2} = 5$ . The range of values for $X_2$ is $[0, 9]$ with each digit being equally likely to be chosen, so $E[X_2] = \frac{1}{10}\frac{9\left(10\right)}{2} = 4.5$ . Thus, $E[S] = 5 + 6(4.5) = \boxed{32}$