What z score in a normal distribution has 33% of all score above it?
Answer: A z score which has 33% of all scores above it, will have 67% of all scores below it.
To find the required z score, we need to find the z value corresponding to probability 0.67.
Using the standard normal table, we have:
[tex]z(0.67)=0.44[/tex]
Therefore, the z score = 0.44 has 33% of all score above it.
Using the normal distribution, it is found that the z-score that has 33% of all scores above it is of Z = 0.44.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The z-score that 33% of all score above it is the 100 - 33 = 67th percentile, which is Z with a p-value of 0.67, so Z = 0.44.
More can be learned about the normal distribution at https://brainly.com/question/24537145
