n=1000, bell-shaped
Mean - 5,000
~950 lie +/-100 of 5000
-4900 to 5100
Required
1) Standard Deviation
2) 97.5th percentile value
3) Z-score for 16th percentile
To find the answers to these questions, we will assume that the distribution of the data follows a normal distribution or bell-shaped curve.
1) Standard Deviation:
The standard deviation can be found using the information provided. We know that roughly 950 out of 1000 data points lie within +/- 100 of the mean (5000). This represents approximately 95% of the data, which falls within two standard deviations (2σ) from the mean in a normal distribution.
Therefore, we can set up the following equation:
2σ = 100
σ = 100 / 2
σ = 50
So, the standard deviation is 50.
2) 97.5th Percentile Value:
To find the 97.5th percentile value, we need to determine the value below which 97.5% of the data falls.
First, calculate the z-score associated with the 97.5th percentile. The z-score formula is:
z = (x - μ) / σ
Where x is the desired percentile value, μ is the mean, and σ is the standard deviation.
We can rewrite the formula to solve for x:
x = (z * σ) + μ
To find the z-score for the 97.5th percentile, we can use a standard normal distribution table or a z-score calculator. The z-score that corresponds to the 97.5th percentile is approximately 1.96.
Using the z-score and the given mean and standard deviation, we can calculate the 97.5th percentile value:
x = (1.96 * 50) + 5000
x = 98 + 5000
x ≈ 5098
So, the 97.5th percentile value is approximately 5098.
3) Z-score for 16th Percentile:
To find the z-score for the 16th percentile, we can use a similar approach as above.
First, we need to find the desired percentile value (x) from the z-score:
x = (z * σ) + μ
Rearranging the equation to solve for z:
z = (x - μ) / σ
Using the given mean and standard deviation, we can calculate the z-score for the 16th percentile:
x = (0.16 * 100) + 5000
x = 16 + 5000
x = 5016
z = (5016 - 5000) / 50
z = 16 / 50
z ≈ 0.32
So, the z-score for the 16th percentile is approximately 0.32.