Final answer:
To find the z-scores less than 2 standard deviations from the mean, you use z-scores of -2 and +2. The Z-table gives areas to the left of these scores, and their difference represents the area between them.
Step-by-step explanation:
The question asks for the z-scores that correspond to less than 2 standard deviations away from the mean in a standard normal distribution. In a standard normal distribution, z-scores of -2 and +2 mark the thresholds of central 95% of the data, according to the empirical rule. To find the area between two z-scores, you look up each z-score in a Z-table to determine the area to the left of each. The area between the z-scores of -2 and +2 can be found by subtracting the area to the left of -2 from the area to the left of +2. According to Z-tables, the area to the left of the z-score -2 is approximately 0.0228 and the area to the left of +2 is approximately 0.9772. The area between these scores, representing the central 95%, is thus 0.9772 - 0.0228, which equals 0.9544, or 95.44% when converted to percentage.
Using the z-table, we can find that the z-score for 2 standard deviations away from the mean is approximately 2.00. To find the area between the z-scores, we can use the z-table to find the area to the left of -2.00 (which is approximately 0.0228) and the area to the left of 2.00 (which is approximately 0.9772). The difference between these areas is approximately 0.9544.