From the given information, we know that the mean and standard deviare are, respectively,
[tex]\begin{gathered} \mu=225 \\ \sigma=10 \end{gathered}[/tex]
Then, we need to use the z score value in order to answer part a and b.
Part a.
In this case, we need to find the z score values for 200 and 220 calories. For the first case, we have
[tex]z=\frac{X-\mu}{\sigma}\Rightarrow z=\frac{200-225}{10}[/tex]
which gives
[tex]z=-2.5[/tex]
For the second case, we have
[tex]z=\frac{X-\mu}{\sigma}\Rightarrow z=\frac{220-225}{10}=-0.5[/tex]
So, we need to find the probability between z= -2.5 and z= -0.5 on the z-table. This is shown the the following picture:
Therefore, the answer for part a is: 0.30233
Part b.
In this case, we need to find the z score value for 190 calories and get the respective probability. Then, the z score is given by
[tex]z=\frac{190-225}{10}=-3.5[/tex]
So, we need to find the following probability
[tex]P(z<-3.5)[/tex]
which is equal to
[tex]P(z<-3.5)=0.00023263[/tex]
as it is shown in the following picture:
Therefore, the answer for part b is: 0.00023263
