Agrocery store will only accept yellow onions that are at least 2.75 in. in diameter. a grower has a crop of onions with diameters that are normally distributed, with a mean diameter of 3.25 in. and a standard deviation of 0.25 in. what percent of the onions will be accepted by the grocery store?

Imagine (or draw) a standard normal curve.  Draw a vertical line through the center of this curve, and mark it 3.25 inches.  This is the mean diameter.  Next, draw vertical lines to represent these sums:  3.25 plus .25, or 3.50 inches; 3.25 less .25, or 3.00 inches, and 3.25 less .50 inches, or 2.75 inches.

Note that 2.75 inches is 2 standard deviations below the mean:  That's 2.75 inches less twice 0.25 inches, or 2.75 = 3.25 inches less 2(0.25) inches.

onions with diameters less than 2.75 are rejected; onions with diameters greater than 2.75 inches are accepted.

Find the area to the LEFT of 2.75 inches.  In other words, find the area to the left of 2 standard deviations below the mean.

I would use the Empirical Rule here:  "68% of data is within 1 standard dev. of the mean; 95% is within 2 standard dev. of the mean.

This tells us that the area to the left of 2.75 inches or -2 standard deviations is 0.025 (half of 0.050).  Subtract this from 1.000; you'll get 0.975.  This means that 97.5% of the onions will prove to be acceptable, whereas 2.5% will be rejected because they are less than 2.75 inches in diameter.

You could also use a table of z-scores to do this:  Determine from the table the area under the std. normal curve to the left of -2 std. dev.  Again, this comes out to 0.025, meaning that the area to the right of -2 std. dev is 0.975.