Find the z scores that separate the middle 59% of the distribution from the area in the tails of the standard normal distribution.
To find the z scores that separate the middle 59% of the distribution from the area in the tails of the standard normal distribution, we can use the Z-table or a statistical calculator.
Step 1: Determine the total area in the tails.
Since the middle 59% of the distribution is being separated, we can calculate the area in the tails as 1 - 0.59 = 0.41.
Step 2: Divide the total area in the tails by 2 to find the area in each tail.
Since the standard normal distribution is symmetric, we divide 0.41 by 2 to get 0.205.
Step 3: Find the z score that corresponds to the area in each tail.
Using the Z-table or a statistical calculator, we can find that the z score for an area of 0.205 in the tails is approximately 0.853.
Therefore, the z scores that separate the middle 59% of the distribution from the area in the tails of the standard normal distribution are approximately -0.853 and 0.853.
To find the z-scores that separate the middle 59% of the distribution from the area in the tails of the standard normal distribution, we need to use the concept of percentiles and the standard normal distribution table (also known as the z-score table or the standard normal distribution curve).
Here's how to do it:
1. Start by understanding the standard normal distribution. It is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. The distribution is symmetrical and bell-shaped.
2. In a standard normal distribution, the area under the curve between any two z-scores represents the percentage of values within that range. For example, the area between z = -1 and z = 1 represents approximately 68% of the values.
3. Since we want to find the z-scores that separate the middle 59% of the distribution from the area in the tails, we need to find the z-scores that represent the middle 59% and subtract them from 1 (to get the area in the tails) and divide by 2 (to split it into two equal halves).
4. To find the z-score that separates the lower tail from the middle 59%, we need to find the cumulative probability that corresponds to (1 - 59%) / 2 = 0.205. This can be done by referring to the z-score table or by using statistical software or calculators.
5. Similarly, to find the z-score that separates the upper tail from the middle 59%, we need to find the cumulative probability that corresponds to (1 + 59%) / 2 = 0.795.
By referring to the z-score table or using statistical software, we can find that the z-score for the lower tail is approximately -0.83 and the z-score for the upper tail is approximately 0.83.
Therefore, the z-scores that separate the middle 59% of the distribution from the area in the tails of the standard normal distribution are approximately -0.83 and 0.83.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability
(± .295) and the Z scores.