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What percentage of scores in a normal distribution fall between a z score of 36 and a z score of -25?
61
What else can I help you with?
What percentage of scores are between 61 and 82?
To determine the percentage of scores between 61 and 82, you would need to know the distribution of the scores (e.g., normal distribution) and the total number of scores. If the data is normally distributed, you can use the mean and standard deviation to find the percentage of scores in that range using a z-score table. Without specific data, it isn't possible to provide an exact percentage.
What percent of the scores are between 63 and 90?
To determine the percentage of scores between 63 and 90, you would need the complete dataset or a statistical summary (like a frequency distribution or histogram) of the scores. By counting the number of scores within that range and dividing by the total number of scores, then multiplying by 100, you can calculate the percentage. Without specific data, it's impossible to provide an exact percentage.
What percentage of the area would the Empirical Rule say is between z -3.00 and z 3.00?
Assuming that you are refering to the standard normal distribution and the z-scores, the answer is 99.73%. If the assumption is incorrect, please resubmit the questionwith more information.
How do you to get the probability in normal distribution?
You calculate the z-scores and then use published tables.
For a normal distribution what z-score value separates the lowest 10 percent of the scores from the rest of the distribution?
-1.28
What percentage of scores fall between 0 and -2 in a normal distribution?
2
What percentage of scores are between 61 and 82?
To determine the percentage of scores between 61 and 82, you would need to know the distribution of the scores (e.g., normal distribution) and the total number of scores. If the data is normally distributed, you can use the mean and standard deviation to find the percentage of scores in that range using a z-score table. Without specific data, it isn't possible to provide an exact percentage.
What percentage of scores in a normal distribution would fall between z-scores of 1 and -2?
3
What percentage of scores fall within -3 and plus 3 standard deviations around the mean in a normal distribution?
99.7% of scores fall within -3 and plus 3 standard deviations around the mean in a normal distribution.
What percentage of normally distributed scores lie under the normal curve?
100%. And that is true for any probability distribution.
How do you find normal distribution of z-scores?
z-scores are distributed according to the standard normal distribution. That is, with the parameters: mean 0 and variance 1.
How would you characterize the distribution of scores in a normal distribution?
They are said to be Normally distributed.
What percent of the scores are between 63 and 90?
To determine the percentage of scores between 63 and 90, you would need the complete dataset or a statistical summary (like a frequency distribution or histogram) of the scores. By counting the number of scores within that range and dividing by the total number of scores, then multiplying by 100, you can calculate the percentage. Without specific data, it's impossible to provide an exact percentage.
When to you use a z scores or t scores?
If the distribution is Gaussian (or Normal) use z-scores. If it is Student's t, then use t-scores.
Intelligence scores follow what kind of distribution?
The IQs of a large enough population can be modeled with a Normal Distribution
What proportion of the scores in a normal distribution is approximately between z -1.16 and z 1.16?
Between z = -1.16 and z = 1.16 is approx 0.7540 (or 75.40 %). Which means ¾ (0.75 or 75%) of the normal distribution lies between approximately -1.16 and 1.16 standard deviations from the mean.
What percentage of the area would the Empirical Rule say is between z -3.00 and z 3.00?
Assuming that you are refering to the standard normal distribution and the z-scores, the answer is 99.73%. If the assumption is incorrect, please resubmit the questionwith more information.
