Final answer:
To find the proportion of the normal distribution that corresponds to z-score values greater than the child's z-score on the science test, we can calculate the z-score for the child's time on the science test using the formula z = (x - μ) / σ. Then, we can use the z-table to find the proportion of the normal distribution that corresponds to the calculated z-score.
Step-by-step explanation:
The proportion of the normal distribution that corresponds to z-score values greater than the child's z-score on the science test can be found using the z-table. Since we want the values that are greater than the child's z-score, we need to find the area under the curve that is above the z-score value. In this case, we can calculate the z-score for the child's time on the science test using the formula z = (x - μ) / σ, where x is the child's time, μ is the population mean, and σ is the population standard deviation.
For the child's time on the science test, we have x = 40 minutes, μ = 31 minutes, and σ = 4 minutes. Plugging these values into the formula, we get z = (40 - 31) / 4 = 2.25. Using the z-table, we can find the proportion of the normal distribution that corresponds to a z-score of 2.25. The table provides the area to the left of a given z-score, so to find the area to the right of the z-score, we subtract the area to the left from 1. Therefore, the proportion of the normal distribution that corresponds to z-score values greater than the child's z-score on the science test is 1 - P(Z < 2.25).