z-score of a value=(that value minus the mean)/(standard deviation). So if a value has a negative z-score, then it is below the mean.
Yes. If a score is below the mean, the z score will be negative.
In a standard normal distribution curve, one half of the area is .5 (or 50%). 0 is the middle value of the z-score. So, for an area of .7704, z must be negative. Also, the area from 0 to z (which is negative) must be equal to .2704. From the normal probability table, this value is -0.74 Therefore, the z-score for the area equals 0.7704 is -0.74
A z score of -1.3 means that the score is located at the negative 1.3 sigma level with respect to the mean.
The z score is -2 or +2!
A negative z score is a value that is less than the mean value.
z-score of a value=(that value minus the mean)/(standard deviation). So if a value has a negative z-score, then it is below the mean.
no, z score can be negative but a probability is a always positive between 0 and 1.
Yes. If a score is below the mean, the z score will be negative.
A negative z-score indicates that the observed value (or statistic) was below the mean. In non-directional tests, a negative z-score is just as likely as a positive one.
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
In a standard normal distribution curve, one half of the area is .5 (or 50%). 0 is the middle value of the z-score. So, for an area of .7704, z must be negative. Also, the area from 0 to z (which is negative) must be equal to .2704. From the normal probability table, this value is -0.74 Therefore, the z-score for the area equals 0.7704 is -0.74
A z-score of 0 means the value is the mean.
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).
z-score of a value=(that value minus the mean)/(standard deviation)
z-score of a value=(that value minus the mean)/(standard deviation)
z-score of a value=(that value minus the mean)/(standard deviation)