How to find area left of a z score
You can find area to the left of a z score (where z is greater than the mean) by using a z-table. The trick to using a z-table is knowing that the numbers in a z-table represent percentages. For example, a value in the table of 0.5000 represents 50% of the area under the curve and a value of .9999 represents 99% of the area under a curve. Once you know how to read the table, finding the area only takes a step or two!
If you are looking for other variations on area, see the index article, area under a normal distribution curve. You’ll find several articles for all different possibilities of areas. For example, finding the area for a value between 0 and any z-score, or an area to the right of a z-score.
How to find area left of a z score: Steps
Step 1: Split your given decimal into two by decimal places. For example, if you’re given 0.46, split that into 0.4 + 0.06.
Step 2: Look up your decimals from Step 1 in the z-table. The z-table below gives the result from looking up 0.4 in the left column and 0.06 in the top row. The intersection (i.e. the area under the curve) is .1772.
Step 3: Add 0.500 to the z-value you just found in step 2.
Note: You’re adding .500 because that’s the 50% of the graph between the mean at zero and the far left of the graph. The above steps only gave you the sliver between 0 and the z-score.
*note on How to find area left of a z score with negative values. The bell curve is symmetrical, so if you are given negative values you can just look up their absolute values. For example, if you are asked for the area of a tail on the left to -0.96, look up the absolute value of -0.96 (0.96).
If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track.Comments are now closed for this post. Need help or want to post a correction? Please post a comment on our Facebook page and I'll do my best to help!