To find the mean from a raw score, z-score, and standard deviation, you can use the formula: ( \text{Raw Score} = \text{Mean} + (z \times \text{Standard Deviation}) ). Rearranging this gives you the mean: ( \text{Mean} = \text{Raw Score} - (z \times \text{Standard Deviation}) ). Simply substitute the values of the raw score, z-score, and standard deviation into this formula to calculate the mean.
Can someone help me find the answer for a sample n=36 with a population mean of of 76 and a mean of 79.4 with a standard deviation of 18?
78
A z-score requires the mean and standard deviation (or standard error). There is, therefore, not enough information to answer the question.
To convert a raw score into a T-score, you first need the mean and standard deviation of the raw scores. The T-score is calculated using the formula: ( T = 50 + 10 \times \frac{(X - \text{Mean})}{\text{SD}} ), where ( X ) is the raw score, Mean is the average of the raw scores, and SD is the standard deviation. This transformation standardizes the score, placing it on a scale where the average is 50 and the standard deviation is 10.
Bob scored 300 or 700.
z-score of a value=(that value minus the mean)/(standard deviation)
score of 92
To get a z-score one needs a standard deviation and a mean as well as the number.
Information is not sufficient to find mean deviation and standard deviation.
A negative Z-Score corresponds to a negative standard deviation, i.e. an observation that is less than the mean, when the standard deviation is normalized so that the standard deviation is zero when the mean is zero.
A z-score cannot help calculate standard deviation. In fact the very point of z-scores is to remove any contribution from the mean or standard deviation.
Can someone help me find the answer for a sample n=36 with a population mean of of 76 and a mean of 79.4 with a standard deviation of 18?
mean= 100 standard deviation= 15 value or x or n = 110 the formula to find the z-value = (value - mean)/standard deviation so, z = 110-100/15 = .6666666 = .6667
No. The standard deviation is not exactly a value but rather how far a score deviates from the mean.
The standard deviation.z-score of a value=(that value minus the mean)/(standard deviation)
78
mean is 218 with a standard deviation of 16