Final answer:
The z-score for a data value of 129 in a normal distribution with a mean of 95 and a standard deviation of 9 is approximately 3.78, indicating the data value is 3.78 standard deviations above the mean.
Step-by-step explanation:
The z-score for a data value indicates how many standard deviations a data point is from the mean. To calculate the z-score, use the formula:
Z = (X - μ) / σ
Where:
- Z is the z-score,
- X is the data value,
- μ (mu) is the mean, and
- σ (sigma) is the standard deviation.
For a data value of 129 with a mean (μ) of 95 and a standard deviation (σ) of 9:
Z = (129 - 95) / 9 = 34 / 9 ≈ 3.78
Thus, a data value of 129 is approximately 3.78 standard deviations above the mean in this normal distribution.