The area between the mean and a Z score of +1.50 is 43.32%. This score is higher than ___ of the score in the distribution.
Assuming a normal distribution, the mean, mode and median have essentially the same value. Median = 50th percentile.
50% + 43.32% = ?
To determine the percentage of scores that are lower than a Z-score of +1.50, we need to subtract the given area of 43.32% from 100%.
100% - 43.32% = 56.68%
Therefore, the score is higher than 56.68% of the scores in the distribution.
To answer this question, we need to understand the concept of Z-scores and their corresponding areas in the standard normal distribution.
A Z-score represents the number of standard deviations a particular value is away from the mean. In other words, it tells us how many standard deviations a data point is above or below the mean in a distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1.
Given that the area between the mean and a Z score of +1.50 is 43.32%, we can determine that this area represents the proportion of scores below the Z score of +1.50.
To find the percentage of scores that are lower than this Z score, we want to subtract the given area (43.32%) from 100%.
100% - 43.32% = 56.68%
Therefore, the Z score of +1.50 is higher than approximately 56.68% of the scores in the distribution.