A raffle uses n tickets, numbered in an arithmetic sequence: the first ticket is numbered 5, and…

Step 1. The variance of $1, 2, \dots, n$. The mean is $\dfrac{n + 1}{2}$. The mean of the squares is $\dfrac{(n + 1)(2n + 1)}{6}$. So $\sigma^2 = \dfrac{(n + 1)(2n + 1)}{6} - \dfrac{(n + 1)^2}{4} = \dfrac{(n + 1)\left[2(2n + 1) - 3(n + 1)\right]}{12} = \dfrac{(n + 1)(n - 1)}{12} = \dfrac{n^2 - 1}{12}$.

Step 2. The ticket numbers. They are $5, 8, 11, \dots$, that is $3k + 2$ for $k = 1, 2, \dots, n$. Adding $2$ does not change the spread, and multiplying by $3$ multiplies the variance by $3^2 = 9$. So the variance is $9 \times \dfrac{n^2 - 1}{12} = \dfrac{3(n^2 - 1)}{4}$.

Step 3. Find $n$. $\dfrac{3(n^2 - 1)}{4} = 90$ gives $n^2 - 1 = 120$, so $n^2 = 121$ and $n = 11$.

Step 4. The mean. The middle ticket is the sixth, numbered $5 + 3 \times 5 = 20$, and the numbers are symmetric about it. So the mean is $20$.

Step 5. The standard deviation. $\sigma = \sqrt{90} = 9.49$.