91Ó°ÊÓ

Probability distribution :

Mean : $\mathrm{μ}X=23.8$

Standard deviation :$\mathrm{σ}X=12.63.$

For every $10$points score one ticket is received it means that

Total number of tickets$\left(T\right)=\frac{Numberofpoints}{10}$

If each data point is divided by the constant, the form of the distribution stays unchanged. Because the lowest bar in the histogram is to the right and the head of higher bars is to the right, the distribution is skewed to the right, and $T$has the same shape as the distribution.

Probability distribution :

Mean :$\mathrm{μ}X=23.8$

Standard deviation :$\mathrm{σ}X=12.63.$

$$\begin{gathered} {\mathrm{μ}}_M=\frac{{\mathrm{μ}}_M}{10} \\ =\frac{23.8}{10} \\ =\mathbf{2}\mathbf{.}\mathbf{38}\mathbf{} \end{gathered}$$

One ticket is given for every ten points earned, implying that

The total number of tickets$\left(T\right)=\frac{Numberofpoints}{10}$

As a result, if all of the data is split by the constant, the distribution's center will be changed in the same way, and the mean will be divided by the same constant $10$.

Probability distribution :

Mean :$\mathrm{μ}X=23.8$

Standard deviation : $\mathrm{σ}X=12.63.$

$$\begin{gathered} {\mathrm{σ}}_m=\frac{{\mathrm{σ}}_X}{10} \\ =\frac{12.63}{10} \\ =\mathbf{1}\mathbf{.}\mathbf{263}\mathbf{} \end{gathered}$$

One ticket is given for every ten points earned, implying that

The total number of tickets $\left(T\right)=\frac{Numberofpoints}{10}$

So, if every piece of data is divided by the constant, the spread of the distribution$T$will be affected in the same way, so the standard deviation is also dived by the same constant $10$, and the number of tickets received on any given roll varies on average by $1.236$from the mean number of tickets.