QUESTION IMAGE
Question
compute the mean, range, and standard deviation for the data items in each of the three samples. then describe one way in which the samples are alike and one way in which they are different.
sample a: 28, 32, 36, 40, 44, 48, 52
sample b: 28, 30, 32, 40, 48, 50, 52
sample c: 28, 28, 28, 40, 52, 52, 52
sample a
mean
40
range
24
standard deviation
8.64 (round to two decimal places as needed)
sample b
mean
range
standard deviation
(round to two decimal places as needed)
Sample B Calculations:
Mean:
Step1: Sum the data values
Data for Sample B: 28, 30, 32, 40, 48, 50, 52
Sum = \( 28 + 30 + 32 + 40 + 48 + 50 + 52 \)
\( = 28 + 30 = 58 \); \( 58 + 32 = 90 \); \( 90 + 40 = 130 \); \( 130 + 48 = 178 \); \( 178 + 50 = 228 \); \( 228 + 52 = 280 \)
Step2: Divide by number of values (n=7)
Mean = \( \frac{280}{7} = 40 \)
Range:
Step1: Identify max and min
Max = 52, Min = 28
Step2: Subtract min from max
Range = \( 52 - 28 = 24 \)
Standard Deviation:
Step1: Find deviations from mean (\( x - \mu \))
Mean (\( \mu \)) = 40
Deviations:
\( 28 - 40 = -12 \); \( 30 - 40 = -10 \); \( 32 - 40 = -8 \); \( 40 - 40 = 0 \); \( 48 - 40 = 8 \); \( 50 - 40 = 10 \); \( 52 - 40 = 12 \)
Step2: Square the deviations (\( (x - \mu)^2 \))
\( (-12)^2 = 144 \); \( (-10)^2 = 100 \); \( (-8)^2 = 64 \); \( 0^2 = 0 \); \( 8^2 = 64 \); \( 10^2 = 100 \); \( 12^2 = 144 \)
Step3: Sum the squared deviations
Sum = \( 144 + 100 + 64 + 0 + 64 + 100 + 144 \)
\( = 144 + 100 = 244 \); \( 244 + 64 = 308 \); \( 308 + 0 = 308 \); \( 308 + 64 = 372 \); \( 372 + 100 = 472 \); \( 472 + 144 = 616 \)
Step4: Divide by \( n - 1 \) (for sample standard deviation)
\( n - 1 = 7 - 1 = 6 \)
Variance = \( \frac{616}{6} \approx 102.67 \)
Step5: Take the square root
Standard Deviation = \( \sqrt{102.67} \approx 10.13 \)
Sample C Calculations:
Mean:
Step1: Sum the data values
Data for Sample C: 28, 28, 28, 40, 52, 52, 52
Sum = \( 28 + 28 + 28 + 40 + 52 + 52 + 52 \)
\( 28 \times 3 = 84 \); \( 52 \times 3 = 156 \); \( 84 + 40 + 156 = 280 \)
Step2: Divide by number of values (n=7)
Mean = \( \frac{280}{7} = 40 \)
Range:
Step1: Identify max and min
Max = 52, Min = 28
Step2: Subtract min from max
Range = \( 52 - 28 = 24 \)
Standard Deviation:
Step1: Find deviations from mean (\( x - \mu \))
Mean (\( \mu \)) = 40
Deviations:
\( 28 - 40 = -12 \) (three times); \( 40 - 40 = 0 \); \( 52 - 40 = 12 \) (three times)
Step2: Square the deviations (\( (x - \mu)^2 \))
\( (-12)^2 = 144 \) (three times); \( 0^2 = 0 \); \( 12^2 = 144 \) (three times)
Step3: Sum the squared deviations
Sum = \( 144 \times 3 + 0 + 144 \times 3 = 432 + 432 = 864 \)
Step4: Divide by \( n - 1 \) (for sample standard deviation)
\( n - 1 = 7 - 1 = 6 \)
Variance = \( \frac{864}{6} = 144 \)
Step5: Take the square root
Standard Deviation = \( \sqrt{144} = 12.00 \)
Summary (All Samples):
| Sample | Mean | Range | Standard Deviation |
|---|---|---|---|
| B | 40 | 24 | 10.13 |
| C | 40 | 24 | 12.00 |
Similarity & Difference:
- Similarity: All samples have the same mean (40) and range (24).
- Difference: The standard deviation increases from Sample A to C, indicating greater variability in data distribution (Sample A has values more spread around the mean, while Sample C has clusters at 28 and 52 with a gap at 40).
Final Answers (Sample B):
- Mean: \( \boldsymbol{40} \)
- Range: \( \boldsymbol{24} \)
- Standard Deviation: \( \boldsymbol{10.13} \) (rounded to two decimals)
Final Answers (Sample C):
- Mean: \( \boldsymbol{40} \)
- Range: \( \boldsymbol{24} \)
- Standard Deviation: \( \boldsymbol{12.00} \)
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Sample B Calculations:
Mean:
Step1: Sum the data values
Data for Sample B: 28, 30, 32, 40, 48, 50, 52
Sum = \( 28 + 30 + 32 + 40 + 48 + 50 + 52 \)
\( = 28 + 30 = 58 \); \( 58 + 32 = 90 \); \( 90 + 40 = 130 \); \( 130 + 48 = 178 \); \( 178 + 50 = 228 \); \( 228 + 52 = 280 \)
Step2: Divide by number of values (n=7)
Mean = \( \frac{280}{7} = 40 \)
Range:
Step1: Identify max and min
Max = 52, Min = 28
Step2: Subtract min from max
Range = \( 52 - 28 = 24 \)
Standard Deviation:
Step1: Find deviations from mean (\( x - \mu \))
Mean (\( \mu \)) = 40
Deviations:
\( 28 - 40 = -12 \); \( 30 - 40 = -10 \); \( 32 - 40 = -8 \); \( 40 - 40 = 0 \); \( 48 - 40 = 8 \); \( 50 - 40 = 10 \); \( 52 - 40 = 12 \)
Step2: Square the deviations (\( (x - \mu)^2 \))
\( (-12)^2 = 144 \); \( (-10)^2 = 100 \); \( (-8)^2 = 64 \); \( 0^2 = 0 \); \( 8^2 = 64 \); \( 10^2 = 100 \); \( 12^2 = 144 \)
Step3: Sum the squared deviations
Sum = \( 144 + 100 + 64 + 0 + 64 + 100 + 144 \)
\( = 144 + 100 = 244 \); \( 244 + 64 = 308 \); \( 308 + 0 = 308 \); \( 308 + 64 = 372 \); \( 372 + 100 = 472 \); \( 472 + 144 = 616 \)
Step4: Divide by \( n - 1 \) (for sample standard deviation)
\( n - 1 = 7 - 1 = 6 \)
Variance = \( \frac{616}{6} \approx 102.67 \)
Step5: Take the square root
Standard Deviation = \( \sqrt{102.67} \approx 10.13 \)
Sample C Calculations:
Mean:
Step1: Sum the data values
Data for Sample C: 28, 28, 28, 40, 52, 52, 52
Sum = \( 28 + 28 + 28 + 40 + 52 + 52 + 52 \)
\( 28 \times 3 = 84 \); \( 52 \times 3 = 156 \); \( 84 + 40 + 156 = 280 \)
Step2: Divide by number of values (n=7)
Mean = \( \frac{280}{7} = 40 \)
Range:
Step1: Identify max and min
Max = 52, Min = 28
Step2: Subtract min from max
Range = \( 52 - 28 = 24 \)
Standard Deviation:
Step1: Find deviations from mean (\( x - \mu \))
Mean (\( \mu \)) = 40
Deviations:
\( 28 - 40 = -12 \) (three times); \( 40 - 40 = 0 \); \( 52 - 40 = 12 \) (three times)
Step2: Square the deviations (\( (x - \mu)^2 \))
\( (-12)^2 = 144 \) (three times); \( 0^2 = 0 \); \( 12^2 = 144 \) (three times)
Step3: Sum the squared deviations
Sum = \( 144 \times 3 + 0 + 144 \times 3 = 432 + 432 = 864 \)
Step4: Divide by \( n - 1 \) (for sample standard deviation)
\( n - 1 = 7 - 1 = 6 \)
Variance = \( \frac{864}{6} = 144 \)
Step5: Take the square root
Standard Deviation = \( \sqrt{144} = 12.00 \)
Summary (All Samples):
| Sample | Mean | Range | Standard Deviation |
|---|---|---|---|
| B | 40 | 24 | 10.13 |
| C | 40 | 24 | 12.00 |
Similarity & Difference:
- Similarity: All samples have the same mean (40) and range (24).
- Difference: The standard deviation increases from Sample A to C, indicating greater variability in data distribution (Sample A has values more spread around the mean, while Sample C has clusters at 28 and 52 with a gap at 40).
Final Answers (Sample B):
- Mean: \( \boldsymbol{40} \)
- Range: \( \boldsymbol{24} \)
- Standard Deviation: \( \boldsymbol{10.13} \) (rounded to two decimals)
Final Answers (Sample C):
- Mean: \( \boldsymbol{40} \)
- Range: \( \boldsymbol{24} \)
- Standard Deviation: \( \boldsymbol{12.00} \)