A distribution of values is normal with a mean of 65.2 and a standard deviation of 7.4.
Find P32, which is the score separating the bottom 32% from the top 68%.
P32 =
Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
http://davidmlane.com/hyperstat/z_table.html
To find P32, we need to use the standard normal distribution (z-score) table.
P32 represents the score separating the bottom 32%. This means that we want to find the z-score that corresponds to the cumulative probability of 0.32.
Using the standard normal distribution table, we can find the z-score for a cumulative probability of 0.32.
In the table, the closest value to 0.32 is 0.3119. This corresponds to a z-score of -0.48.
Now, we can convert the z-score back to the original distribution by using the formula:
z = (x - mean) / standard deviation
Rearranging the formula, we can solve for x:
x = (z * standard deviation) + mean
Plugging in the values, we get:
x = (-0.48 * 7.4) + 65.2
x = -3.552 + 65.2
x = 61.648
Therefore, P32 is approximately 61.6.
To find P32, the score separating the bottom 32% from the top 68%, we need to use the standard normal distribution.
The standard normal distribution has a mean of 0 and a standard deviation of 1. In order to use it, we need to convert the given values to z-scores.
To convert a value x to a z-score, we use the formula:
z = (x - mean) / standard deviation
Given that the mean of the distribution is 65.2 and the standard deviation is 7.4, we can calculate the z-score as:
z = (32nd percentile score - mean) / standard deviation
We want to find the score separating the bottom 32%, so the 32nd percentile score is the value we are looking for. In the standard normal distribution, the z-score that corresponds to a percentile can be found using a Z-table or a calculator.
Using a Z-table, we can find the z-score that corresponds to a percentile of 32%.
The Z-table shows that a percentile of 32% corresponds to a z-score of approximately -0.405.
Now, we can use this z-score to find the score separating the bottom 32% from the top 68% by rearranging the z-score formula:
32nd percentile score = (z * standard deviation) + mean
Plugging in the values, we get:
32nd percentile score = (-0.405 * 7.4) + 65.2
Simplifying this expression, we find:
32nd percentile score ≈ 62.68
Therefore, P32, the score separating the bottom 32% from the top 68%, is approximately 62.7 (rounded to 1 decimal place).
How to do this on your own.
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability (P=.32) and its Z score.
Z = (score-mean)/SD
Insert data to solve.