If the average (xbar) on an assignment in a class was 82% with a standard deviation (s) of 4.8, what is the z-score for someone scoring 75% (round to the one hundredths place or two decimal places)?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
82-75= 7/4.8=1.45
To find the z-score for someone scoring 75%, we can use the formula:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the individual score (75% in this case)
- μ is the mean (average) score (82% in this case)
- σ is the standard deviation (4.8 in this case)
Plugging in the values, we get:
z = (75 - 82) / 4.8
Calculating this expression, we have:
z = -7 / 4.8
z ≈ -1.46
Thus, the z-score for someone scoring 75% is approximately -1.46 (rounded to two decimal places).