Answer:
Center = 15
Spread = 0.599
Shape = Normal
Explanation:
The provided information is:
![\mu_1=38 \ \ \ \ \ \ \ \ \mu_2 = 23\\\sigma_1 =2 \ \ \ \ \ \ \ \ \ \sigma_2=4\\n_1=45 \ \ \ \ \ \ \ \ n_2 =59](https://img.qammunity.org/2021/formulas/mathematics/college/e1ga3irmezon8lzjhfbsylyz15hvmj5p1g.png)
Thus the center(mean) of the distribution is:
![\begin{aligned}Mean &= \mu_1-\mu_2\\&=38-23\\&=15\end{aligned}](https://img.qammunity.org/2021/formulas/mathematics/college/8f89py7x3prx3vllsg3t8xz5dwrhg5gtn7.png)
The spread (standard deviation) of the distribution is:
![\begin{aligned}\textrm{Standard deviation}&=\sqrt{(\sigma_1^2)/(n_1)+(\sigma_2^2)/(n_2)}\\&=\sqrt{(4)/(45)+(16)/(59)}\\&=0.599\end{aligned}](https://img.qammunity.org/2021/formulas/mathematics/college/atz4yef1a6ps1lcogww0fl3b585178cid0.png)
The shape of the distribution is also normally distributed.