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compute the mean, range, and standard deviation for the data items in e…

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)

Explanation:

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):
SampleMeanRangeStandard Deviation
B402410.13
C402412.00
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: \( \boldsymbol{12.00} \)

Answer:

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):
SampleMeanRangeStandard Deviation
B402410.13
C402412.00
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: \( \boldsymbol{12.00} \)