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.
A z-score of 0 means the value is the mean.
The z-score must be 1.87: the probability cannot have that value!
Z Score is (x-mu)/sigma. The Z-Score allows you to go to a standard normal distribution chart and to determine probabilities or numerical values.
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.
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.
Yes a Z score can be 5.
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
Find the Z score that correspond to P25
A z-score is a linear transformation. There is nothing to "prove".
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).