Question

find the standard deviation for the given sample data. round your answer to one more decimal place than is present in the original data. 22.6 16.1 36.1 36.0 23.8 20.3 a. 4347.7 b. 3999.0 c. 36.1 d. 8.35

Explanation:

Step1: Identify the data set

The data set is \( 22.6, 16.1, 36.1, 36.0, 23.8, 20.3 \) (assuming there are 6 data points as listed).

Step2: Calculate the mean (\(\bar{x}\))

First, sum the data points:
\( 22.6 + 16.1 + 36.1 + 36.0 + 23.8 + 20.3 = 154.9 \)
Then, divide by the number of data points (\(n = 6\)):
\( \bar{x} = \frac{154.9}{6} \approx 25.8167 \)

Step3: Calculate the squared differences from the mean

For each data point \( x_i \), calculate \( (x_i - \bar{x})^2 \):

• \( (22.6 - 25.8167)^2 \approx (-3.2167)^2 \approx 10.347 \)
• \( (16.1 - 25.8167)^2 \approx (-9.7167)^2 \approx 94.414 \)
• \( (36.1 - 25.8167)^2 \approx (10.2833)^2 \approx 105.746 \)
• \( (36.0 - 25.8167)^2 \approx (10.1833)^2 \approx 103.700 \)
• \( (23.8 - 25.8167)^2 \approx (-2.0167)^2 \approx 4.067 \)
• \( (20.3 - 25.8167)^2 \approx (-5.5167)^2 \approx 30.434 \)

Step4: Sum the squared differences

Sum these values:
\( 10.347 + 94.414 + 105.746 + 103.700 + 4.067 + 30.434 \approx 348.708 \)

Step5: Calculate the sample variance (\(s^2\))

Divide the sum of squared differences by \( n - 1 = 5 \):
\( s^2 = \frac{348.708}{5} \approx 69.7416 \)

Step6: Calculate the sample standard deviation (\(s\))

Take the square root of the variance:
\( s = \sqrt{69.7416} \approx 8.35 \) (rounded to two decimal places, but we need to round to one more decimal than original data. Original data has one decimal place, so we round to two decimal places? Wait, original data: 22.6 (1 decimal), 16.1 (1), 36.1 (1), 36.0 (1), 23.8 (1), 20.3 (1). So we need to round to two decimal places? Wait, the problem says "round your answer to one more decimal place than is present in the original data". Original data has one decimal place, so we round to two decimal places. Wait, but let's check the calculation again. Wait, maybe I made a mistake in the number of data points. Wait, the data listed is 22.6, 16.1, 36.1, 36.0, 23.8, 20.3 – that's 6 data points. Wait, but let's recalculate the mean: 22.6 +16.1=38.7; 38.7+36.1=74.8; 74.8+36.0=110.8; 110.8+23.8=134.6; 134.6+20.3=154.9. Mean is 154.9/6 ≈25.8167. Then squared differences:

22.6 -25.8167 = -3.2167, squared: ~10.347

16.1 -25.8167 = -9.7167, squared: ~94.414

36.1 -25.8167 = 10.2833, squared: ~105.746

36.0 -25.8167 = 10.1833, squared: ~103.700

23.8 -25.8167 = -2.0167, squared: ~4.067

20.3 -25.8167 = -5.5167, squared: ~30.434

Sum: 10.347+94.414=104.761; +105.746=210.507; +103.700=314.207; +4.067=318.274; +30.434=348.708. Variance: 348.708/5=69.7416. Standard deviation: sqrt(69.7416)≈8.35. So the answer is D. 8.35.

Answer:

D. 8.35