Could you guys help me solve this problem i really really need some help.. thanks
the tensile strength of a certain metal component is normally distributed with a mean of 10,000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. measurements are recorded to the nearest 50 kilograms per square centimeter. if specifications require that all components have tensile strength between 9800 and 10200 kilograms per square centimeter inclusive, what proportion of pieces would you expect to scrap?

Have you made no attempt at this yourself? Your normal distribution has mean 10000 and standard deviation 100 so that $\displaystyle z= \frac{x- 10000}{100}$ 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 $\displaystyle z=\frac{9750- 10000}{100}= \frac{-250}{100}= -2.5$ 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.