Have you made no attempt at this yourself? Your normal distribution has mean 10000 and standard deviation 100 so that $z= \frac{x- 10000}{100}$ has the "standard" normal distribution. Since all "measurements are recorded to the nearest 50 kilograms" you want to find the proportion (probability) that x is between 9800- 50= 9750 and 10200+ 50= 10250. Those correspond to $z=\frac{9750- 10000}{100}= \frac{-250}{100}= -2.5$ to [tex]z= \frac{10250- 10000}{100}= \frac{250}{100}= 2.5[/itex]. Use a table of the normal distribution to find the proportion between -2.5 and 2.5.