Have you made no attempt at this yourself? Your normal distribution has mean 10000 and standard deviation 100 so that 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 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.