Answer:
Center = 15
Spread = 0.599
Shape = Normal
Step-by-step explanation:
The provided information is:
[tex]\mu_1=38 \ \ \ \ \ \ \ \ \mu_2 = 23\\\sigma_1 =2 \ \ \ \ \ \ \ \ \ \sigma_2=4\\n_1=45 \ \ \ \ \ \ \ \ n_2 =59[/tex]
Thus the center(mean) of the distribution is:
[tex]\begin{aligned}Mean &= \mu_1-\mu_2\\&=38-23\\&=15\end{aligned}[/tex]
The spread (standard deviation) of the distribution is:
[tex]\begin{aligned}\textrm{Standard deviation}&=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\&=\sqrt{\frac{4}{45}+\frac{16}{59}}\\&=0.599\end{aligned}[/tex]
The shape of the distribution is also normally distributed.
