For the following data set, calculate the percentage of data points that fall within one standard deviation of the mean, and compare the result to the expected percentage of a normal distribution.
{50, 46, 54, 51, 29, 52, 48, 54, 47,
A. 80%; This percentage is lower than expected in a normal distribution.
B. 80%; This percentage is higher than expected in a normal distribution.
C. 90%; This percentage is lower than expected in a normal distribution.
D. 90%; This percentage is higher than expected in a normal distribution
To calculate the percentage of data points that fall within one standard deviation of the mean, we first need to calculate the mean and standard deviation of the data set.
Step 1: Calculate the mean:
Mean = (50 + 46 + 54 + 51 + 29 + 52 + 48 + 54 + 47) / 9 = 481 / 9 ≈ 53.4
Step 2: Calculate the standard deviation:
For simplicity, we'll use the population standard deviation formula since we have the entire data set. The formula is:
Standard Deviation = √((Σ(x - µ)^2) / N)
where Σ is the summation symbol, x is each individual data point, µ is the mean, and N is the total number of data points.
Using this formula, we can calculate the standard deviation as follows:
Standard Deviation = √(((50 - 53.4)^2 + (46 - 53.4)^2 + ... + (47 - 53.4)^2) / 9)
≈ √((17.36 + 52.96 + ... + 35.64) / 9)
≈ √(434.24 / 9)
≈ √48.25
≈ 6.94
Step 3: Calculate the percentage of data points within one standard deviation of the mean:
To be within one standard deviation of the mean, the data points must fall within the range of mean ± standard deviation.
Lower Bound = mean - standard deviation = 53.4 - 6.94 ≈ 46.46
Upper Bound = mean + standard deviation = 53.4 + 6.94 ≈ 60.34
Now, out of the 9 data points, we can see that 7 of them (50, 46, 54, 51, 52, 48, 54) fall within the range of 46.46 to 60.34.
Percentage within one standard deviation = (7 / 9) * 100 ≈ 77.8%
The correct answer is not listed in the options, but the calculated percentage (77.8%) is lower than the expected percentage of a normal distribution (which is approximately 68% within one standard deviation).