Question

find the standard deviation for the given sample data. round your answer to one more decimal place than the original data. 18 18 18 9 15 5 10 5 15 a. 5.4 b. 5.8 c. 1.6 d. 5.1

Explanation:

Step1: List the data points

The data set is: \( 18, 18, 18, 18, 9, 15, 5, 10, 5, 15 \)
First, find the number of data points \( n \). Counting them, we have \( n = 10 \).

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

The mean is calculated as \(\bar{x} = \frac{\sum_{i = 1}^{n} x_i}{n}\)
Sum of the data points:
\( 18 + 18 + 18 + 18 + 9 + 15 + 5 + 10 + 5 + 15 \)
\( = (18\times4)+9 + 15\times2+5\times2 + 10 \)
\( = 72+9 + 30+10 + 10 \)
\( = 72 + 9=81; 81+30 = 111; 111+10 = 121; 121 + 10=131 \)
Wait, let's recalculate the sum:
\( 18+18 = 36; 36+18 = 54; 54+18 = 72; 72+9 = 81; 81+15 = 96; 96+5 = 101; 101+10 = 111; 111+5 = 116; 116+15 = 131 \)
So, \(\bar{x}=\frac{131}{10}=13.1\)

Step3: Calculate the squared differences from the mean

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

• For \( x = 18 \): \( (18 - 13.1)^2=(4.9)^2 = 24.01 \) (there are 4 such points, so total for these: \( 4\times24.01 = 96.04 \))
• For \( x = 9 \): \( (9 - 13.1)^2=(-4.1)^2 = 16.81 \)
• For \( x = 15 \): \( (15 - 13.1)^2=(1.9)^2 = 3.61 \) (there are 2 such points, so total: \( 2\times3.61 = 7.22 \))
• For \( x = 5 \): \( (5 - 13.1)^2=(-8.1)^2 = 65.61 \) (there are 2 such points, so total: \( 2\times65.61 = 131.22 \))
• For \( x = 10 \): \( (10 - 13.1)^2=(-3.1)^2 = 9.61 \)

Step4: Sum the squared differences

Sum of squared differences \( S = 96.04+16.81 + 7.22+131.22 + 9.61 \)
\( 96.04+16.81 = 112.85; 112.85+7.22 = 120.07; 120.07+131.22 = 251.29; 251.29+9.61 = 260.9 \)

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

Sample variance is \( s^2=\frac{S}{n - 1}\) (since it's sample data)
\( n-1 = 9 \), so \( s^2=\frac{260.9}{9}\approx28.9889 \)

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

Sample standard deviation is the square root of the sample variance: \( s=\sqrt{s^2}=\sqrt{28.9889}\approx5.4 \) (rounded to one more decimal place than the original data, original data has 0 decimal places, so we round to 1 decimal place? Wait, no, the problem says "Round your answer to one more decimal place than the original data." Original data points are integers (0 decimal places), so we round to 1 decimal place? Wait, but let's check the calculation again. Wait, maybe I made a mistake in the sum of data points. Let's re - calculate the sum:

Data points: 18, 18, 18, 18, 9, 15, 5, 10, 5, 15

Let's add them one by one:

18 + 18 = 36

36+18 = 54

54+18 = 72

72+9 = 81

81+15 = 96

96+5 = 101

101+10 = 111

111+5 = 116

116+15 = 131. So the sum is correct. Mean is 13.1.

Now, recalculate the squared differences:

For 18: \( (18 - 13.1)=4.9 \), \( 4.9^2 = 24.01 \), 4 times: \( 4\times24.01 = 96.04 \)

For 9: \( (9 - 13.1)=- 4.1 \), \( (-4.1)^2 = 16.81 \)

For 15: \( (15 - 13.1)=1.9 \), \( 1.9^2 = 3.61 \), 2 times: \( 2\times3.61 = 7.22 \)

For 5: \( (5 - 13.1)=-8.1 \), \( (-8.1)^2 = 65.61 \), 2 times: \( 2\times65.61 = 131.22 \)

For 10: \( (10 - 13.1)=-3.1 \), \( (-3.1)^2 = 9.61 \)

Sum of squared differences: \( 96.04+16.81+7.22+131.22 + 9.61=96.04+16.81 = 112.85; 112.85+7.22 = 120.07; 120.07+131.22 = 251.29; 251.29+9.61 = 260.9 \). Correct.

Sample variance: \( s^2=\frac{260.9}{9}\approx28.9889 \)

Sample standard deviation: \( s=\sqrt{28.9889}\approx5.4 \) (since \(\sqrt{28.9889}\approx5.384\), which rounds to 5.4 when rounded to one decimal place, and the original data has 0 decimal places, so we round to 1 decimal place, which is one more than 0)

Answer:

A. 5.4