True or False
For any normal distribution, the proportion corresponding to scores greater than z = +1.00 is exactly equal to the proportion corresponding to scores less than z = 1.00.
I'm sorry that this question was posted twice I was having problems with my internet due to the snow storms. I would still like to have help answering this question.
Assuming that your square is a negative sign, it would be true.
To determine whether the statement is true or false, we need to understand the concept of a normal distribution and how z-scores are related to it.
A normal distribution is a bell-shaped probability distribution that is characterized by its mean (μ) and standard deviation (σ). It is symmetric around the mean and follows a specific pattern.
A z-score, also known as a standard score, measures the number of standard deviations a specific value is from the mean of a normal distribution. It allows us to compare values from different normal distributions.
Now, let's address the statement. For any normal distribution, the proportion corresponding to scores greater than z = +1.00 is not exactly equal to the proportion corresponding to scores less than z = -1.00. This is because the normal distribution is symmetric around the mean.
In a standard normal distribution (mean = 0, standard deviation = 1), approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations. This means that approximately 34% of the data falls to the right of z = +1.00 and approximately 34% falls to the left of z = -1.00.
However, for non-standard normal distributions (with a mean other than 0 and a non-unit standard deviation), the proportions will differ depending on the specific distribution's parameters.
To summarize, the statement is false. The proportion corresponding to scores greater than z = +1.00 is not exactly equal to the proportion corresponding to scores less than z = -1.00 in any general normal distribution. The specific proportions depend on the mean and standard deviation of the distribution.