Question

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 {70, 85, 84, 83, 96, 74, 66, 99, 84, 81} (1 point) 50%; this percentage is lower than expected in a normal distribution. 60%; this percentage is lower than expected in a normal distribution. 70%; this percentage is close to the expected percentage in a normal distribution. 80%; this percentage is higher than expected in a normal distribution.

Explanation:

Step1: Calculate the mean

Data set: {70,85,84,83,96,74,66,99,84,81}, n=10
Mean $\mu = \frac{70+85+84+83+96+74+66+99+84+81}{10} = \frac{822}{10} = 82.2$

Step2: Calculate sample standard deviation

Squared deviations:
$(70-82.2)^2=148.84$, $(85-82.2)^2=7.84$, $(84-82.2)^2=3.24$, $(83-82.2)^2=0.64$, $(96-82.2)^2=190.44$,
$(74-82.2)^2=67.24$, $(66-82.2)^2=262.44$, $(99-82.2)^2=282.24$, $(84-82.2)^2=3.24$, $(81-82.2)^2=1.44$
Sum of squared deviations = $148.84+7.84+3.24+0.64+190.44+67.24+262.44+282.24+3.24+1.44=967.6$
Variance $s^2 = \frac{967.6}{10-1} \approx 107.51$
Standard deviation $s \approx \sqrt{107.51} \approx 10.37$

Step3: Find range within 1 SD

Range: $\mu - s \approx 82.2 - 10.37 = 71.83$; $\mu + s \approx 82.2 + 10.37 = 92.57$
Count data points in [71.83,92.57]: 74,81,83,84,84,85 → 6 points

Step4: Calculate percentage

Percentage = $\frac{6}{10} \times 100\% = 60\%$
Normal distribution expects ~68%, so 60% is lower.

Answer:

B. 60%; This percentage is lower than expected in a normal distribution.