91Ó°ÊÓ

The number of tickets is the number of points scored divided by$10$

$T=\frac{1}{10}X$

The probabilities for $T$are the same as the probabilities for $X$.

Create a probability histogram

The width of the bars has to be equal and the height has to be equal to probability.

Given

${\mathrm{μ}}_X=23.8$

${\mathrm{σ}}_X=12.63$

The number of tickets is the number of points scored divided by$10$

$T=\frac{1}{10}X$

The probabilities for $T$are the same as the probabilities for $X$.

Properties mean and standard deviation:

${\mathrm{μ}}_{aX+b}=a{\mathrm{μ}}_X+b$

${\mathrm{σ}}_{aX+b}=a{\mathrm{σ}}_X$

Then we obtain for $T=\frac{1}{10}X$:

localid="1649909964609" $$\begin{gathered} {\mathrm{μ}}_T={\mathrm{μ}}_{1/10*X} \\ =\frac{1}{10}{\mathrm{μ}}_X \\ =\frac{1}{10}(23.8) \\ =2.38 \end{gathered}$$.

Given

${\mathrm{μ}}_X=23.8$

${\mathrm{σ}}_X=12.63$

The number of tickets is the number of points scored divided by$10$

$T=\frac{1}{10}X$

The probabilities for $T$are the same as the probabilities for $X$.

Properties mean and standard deviation:

${\mathrm{μ}}_{aX+b}=a{\mathrm{μ}}_X+b$

${\mathrm{σ}}_{aX+b}=a{\mathrm{σ}}_X$

Then we obtain for $T=\frac{1}{10}X$:

localid="1649910001357" $$\begin{gathered} {\mathrm{μ}}_T={\mathrm{μ}}_{1/10*X} \\ =\frac{1}{10}{\mathrm{μ}}_X \\ =\frac{1}{10}(23.8) \\ =2.38 \end{gathered}$$

localid="1649910014351" $$\begin{gathered} {\mathrm{σ}}_T={\mathrm{σ}}_{1/10*X} \\ =\frac{1}{10}{\mathrm{σ}}_X \\ =\frac{1}{10}(12.63) \\ =1.263 \end{gathered}$$

The number of tickets varies, on average, about $1.263$ticket about the mean of $2.38$tickets.