If a normally distributed random variable X has mean m and standard deviation s, then z = (X - m)/s
T-score is used when you don't have the population standard deviation and must use the sample standard deviation as a substitute.
Mean μ = 63.3 Standard deviation σ = 3.82 Standard error σ / √ n = 3.82 / √ 19 = 0.8763681 z = (xbar - μ) / (σ / √ n ) z = (61.6-63.3) / 0.876368 z = -1.9398
The absolute value of the standard score becomes smaller.
78
27
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.
You need the mean and standard deviation in order to calculate the z-score. Neither are given.
If it is possible to assume normality, simply convert the desired score to a z-score, and look up the probability for that.
score of 92
If a normally distributed random variable X has mean m and standard deviation s, then z = (X - m)/s
T-score is used when you don't have the population standard deviation and must use the sample standard deviation as a substitute.
zero
The standardised score decreases.
When you don't have the population standard deviation, but do have the sample standard deviation. The Z score will be better to do as long as it is possible to do it.
No, it is called the absolute deviation.
standard deviation