No it is not.
No. If the underlying distribution is approximately Normal then 1.4 is not at all unusual.
z score = (test score - mean score)/SD z score = (87-81.1)/11.06z score = 5.9/11.06z score = .533You can use a z-score chart to calculate the probability from there.
The z-score must be 1.87: the probability cannot have that value!
A z-score of 0 means the value is the mean.
If a score ( s ) is equal to the mean of a dataset, its z-score will be 0. The z-score is calculated using the formula ( z = \frac{s - \mu}{\sigma} ), where ( \mu ) is the mean and ( \sigma ) is the standard deviation. Since ( s ) equals ( \mu ), the numerator becomes zero, resulting in a z-score of 0. This indicates that the score is exactly at the average of the dataset.
No. If the underlying distribution is approximately Normal then 1.4 is not at all unusual.
It shows us by telling us how far away it is from the mean.
The Z-score is just the score. The Z-test uses the Z-score to compare to the critical value. That is then used to establish if the null hypothesis is refused.
what is the z score for 0.75
z score = (test score - mean score)/SD z score = (87-81.1)/11.06z score = 5.9/11.06z score = .533You can use a z-score chart to calculate the probability from there.
Yes a Z score can be 5.
It would depend. If my IQ had a z-score of 4 I would be thrilled because it would imply that my IQ was higher than about 99.997% of the population. Of course, on the other hand, I'd be skeptical of that because I've never given any indication of being that smart. I suppose that's what you're getting at: a z-score this high is darned unusual.
If the Z Score of a test is equal to zero then the raw score of the test is equal to the mean. Z Score = (Raw Score - Mean Score) / Standard Deviation
A z-score is a linear transformation. There is nothing to "prove".
Find the Z score that correspond to P25
Assume the z-score is relative to zero score. In simple terms, assume that we have 0 < z < z0, where z0 is the arbitrary value. Then, a negative z-score can be greater than a positive z-score (yes). How? Determine the probability of P(-2 < z < 0) and P(0 < z < 1). Then, by checking the z-value table, you should get: P(-2 < z < 0) ≈ 0.47725 P(0 < z < 1) ≈ 0.341345
Let z be positive so that -z is the negative z score for which you want the probability. Pr(Z < -z) = Pr(Z > z) = 1 - Pr(Z < z).