Question

the three data sets have the same mean and range, but is the variation the same? prove your answer by computing the standard deviation. assume the data were obtained from samples. (a) 3, 7, 9, 11, 13, 13, 14 (b) 3, 8, 8, 11, 11, 13, 15 (c) 3, 6, 9, 10, 12, 14, 15

Explanation:

Step 1: Calculate the mean for each data set

For a data set \( x_1, x_2, \dots, x_n \), the mean \( \bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n} \)

Data set (a): \( 3,7,9,11,13,13,14 \)

\( n = 7 \)
\( \sum x=(3 + 7+9 + 11+13+13+14)=60 \)
\( \bar{x}_a=\frac{60}{7}\approx8.571 \) (Wait, no, 3+7=10, 10+9=19, 19+11=30, 30+13=43, 43+13=56, 56+14=70. Oh, I made a mistake. \( \sum x = 3+7 + 9+11+13+13+14=70 \), so \( \bar{x}_a=\frac{70}{7} = 10 \)

Data set (b): \( 3,8,8,11,11,13,15 \)

\( \sum x=3 + 8+8+11+11+13+15=69 \)? Wait, 3+8=11, 11+8=19, 19+11=30, 30+11=41, 41+13=54, 54+15=69? No, 3+8=11, 11+8=19, 19+11=30, 30+11=41, 41+13=54, 54+15=69? Wait, no, 3 + 8+8+11+11+13+15: 3+8=11, +8=19, +11=30, +11=41, +13=54, +15=69. Then \( \bar{x}_b=\frac{69}{7}\approx9.857 \)? Wait, no, the problem says the three data sets have the same mean. So I must have miscalculated. Let's recalculate:

Data set (a): 3,7,9,11,13,13,14. Sum: 3+7=10, +9=19, +11=30, +13=43, +13=56, +14=70. Mean: 70/7 = 10.

Data set (b): 3,8,8,11,11,13,15. Sum: 3+8=11, +8=19, +11=30, +11=41, +13=54, +15=69. Wait, that's not 70. Wait, maybe I misread the data. Let me check again. Maybe data set (b) is 3,8,8,11,11,13,16? No, the original problem: (b) 3, 8, 8, 11, 11, 13, 15. Wait, maybe the problem statement has a typo, but according to the problem, they have the same mean. So perhaps I made a mistake. Let's check data set (c): 3,6,9,10,12,14,15. Sum: 3+6=9, +9=18, +10=28, +12=40, +14=54, +15=69. No, that's also 69. Wait, the problem says "the three data sets have the same mean and range". So maybe the data is different. Wait, maybe data set (a) is 3,7,9,11,13,13,14 (sum 70, mean 10), data set (b) 3,8,8,11,11,13,16 (sum 3+8+8+11+11+13+16=70, mean 10), data set (c) 3,6,9,10,12,14,16 (sum 3+6+9+10+12+14+16=70, mean 10). Maybe a typo in the original problem. But assuming the problem is correct, and we proceed with the formula for sample standard deviation: \( s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}} \)

Let's assume the mean \( \bar{x}=10 \) for all (since the problem says same mean). Let's recalculate the sums:

Data set (a): \( x_i: 3,7,9,11,13,13,14 \)

\( (x_i - \bar{x})^2 \):
\( (3 - 10)^2=49 \)
\( (7 - 10)^2=9 \)
\( (9 - 10)^2=1 \)
\( (11 - 10)^2=1 \)
\( (13 - 10)^2=9 \)
\( (13 - 10)^2=9 \)
\( (14 - 10)^2=16 \)
Sum of squares: \( 49+9 + 1+1+9+9+16=94 \)
Sample standard deviation \( s_a=\sqrt{\frac{94}{7 - 1}}=\sqrt{\frac{94}{6}}\approx\sqrt{15.6667}\approx3.958 \)

Data set (b): \( x_i: 3,8,8,11,11,13,15 \)

Assume mean \( \bar{x}=10 \) (even if sum was miscalculated, let's use the problem's statement)
\( (x_i - \bar{x})^2 \):
\( (3 - 10)^2=49 \)
\( (8 - 10)^2=4 \)
\( (8 - 10)^2=4 \)
\( (11 - 10)^2=1 \)
\( (11 - 10)^2=1 \)
\( (13 - 10)^2=9 \)
\( (15 - 10)^2=25 \)
Sum of squares: \( 49+4 + 4+1+1+9+25=93 \)
Sample standard deviation \( s_b=\sqrt{\frac{93}{6}}=\sqrt{15.5}\approx3.937 \) Wait, no, if mean is 10, sum of data should be 70. Let's recalculate data set (b) sum: 3+8+8+11+11+13+16=70 (so 15 is a typo, should be 16). Then (16 - 10)^2=36. Then sum of squares: 49+4+4+1+1+9+36=104. Then \( s_b=\sqrt{\frac{104}{6}}\approx\sqrt{17.333}\approx4.163 \). This is getting confusing. Alternatively, maybe the data is:

Data set (a): 3,7,9,11,13,13,14 (sum 70, mean 10)

Data set (b): 3,8,8,11,11,13,16 (sum 70, mean 10)

Data set (c): 3,6,9,10,12,14,16 (sum 70, mean 10)

But perhaps a better approach: the standard deviation measures the spread. Data set (b) has more values clustered around the mean? Wait, no. Wait, let's calculate correctly with the given data (assuming the problem's "same mean" is correct, so maybe my initial sum was wrong).

Wait, let's recalculate data set (a) sum: 3+7=10, +9=19, +11=30, +13=43, +13=56, +14=70. Mean 10. Correct.

Data set (b): 3+8=11, +8=19, +11=30, +11=41, +13=54, +15=69. Mean 69/7≈9.857. Not 10. So there must be a typo. Maybe data set (b) is 3,8,8,11,11,13,16 (sum 70, mean 10). Data set (c): 3,6,9,10,12,14,16 (sum 70, mean 10). Alternatively, maybe the original data is:

(a) 3, 7, 9, 11, 13, 13, 14 (n=7, sum=70, mean=10)

(b) 3, 8, 8, 11, 11, 13, 15 (n=7, sum=3+8+8+11+11+13+15=69, mean=69/7≈9.857) – this contradicts the problem statement. So perhaps the problem has a mistake, but we proceed with the formula.

Alternatively, maybe the range is same. Range = max - min.

For (a): max=14, min=3, range=11.

For (b): max=15, min=3, range=12. No, that's not same. So definitely a typo. Let's assume that the data is correct as per the problem, and maybe the mean is approximately same. Let's calculate the sample standard deviation for each:

Formula for sample standard deviation: \( s=\sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \), where \( \bar{x}=\frac{\sum x_i}{n} \)

Data set (a):

\( n = 7 \)
\( \sum x_i=3 + 7+9 + 11+13+13+14=70 \)
\( \bar{x}_a=\frac{70}{7}=10 \)
\( \sum (x_i - \bar{x}_a)^2=(3 - 10)^2+(7 - 10)^2+(9 - 10)^2+(11 - 10)^2+(13 - 10)^2+(13 - 10)^2+(14 - 10)^2 \)
\( =49 + 9+1 + 1+9+9+16=94 \)
\( s_a=\sqrt{\frac{94}{6}}\approx\sqrt{15.6667}\approx3.958 \)

Data set (b):

\( n = 7 \)
\( \sum x_i=3 + 8+8 + 11+11+13+15=69 \)
\( \bar{x}_b=\frac{69}{7}\approx9.857 \)
\( \sum (x_i - \bar{x}_b)^2=(3 - 9.857)^2+(8 - 9.857)^2+(8 - 9.857)^2+(11 - 9.857)^2+(11 - 9.857)^2+(13 - 9.857)^2+(15 - 9.857)^2 \)
\( \approx(-6.857)^2+(-1.857)^2+(-1.857)^2+(1.143)^2+(1.143)^2+(3.143)^2+(5.143)^2 \)
\( \approx47.01 + 3.45 + 3.45 + 1.31 + 1.31 + 9.88 + 26.45 \approx92.86 \)
\( s_b=\sqrt{\frac{92.86}{6}}\approx\sqrt{15.477}\approx3.934 \)

Data set (c):

\( n = 7 \)
\( \sum x_i=3 + 6+9 + 10+12+14+15=69 \)
\( \bar{x}_c=\frac{69}{7}\approx9.857 \)
\( \sum (x_i - \bar{x}_c)^2=(3 - 9.857)^2+(6 - 9.857)^2+(9 - 9.857)^2+(10 - 9.857)^2+(12 - 9.857)^2+(14 - 9.857)^2+(15 - 9.857)^2 \)
\( \approx(-6.857)^2+(-3.857)^2+(-0.857)^2+(0.143)^2+(2.143)^2+(4.143)^2+(5.143)^2 \)
\( \approx47.01 + 14.87 + 0.73 + 0.02 + 4.59 + 17.16 + 26.45 \approx110.83 \)
\( s_c=\sqrt{\frac{110.83}{6}}\approx\sqrt{18.472}\approx4.298 \)

Wait, but the problem says "the three data sets have the same mean and range". So maybe the data is different. Let's assume that the mean is 10 for all, so data set (b) should have sum 70, so 15 is replaced with 16 (3+8+8+11+11+13+16=70), and data set (c) 3+6+9+10+12+14+16=70. Then:

Data set (b) corrected:

\( \sum x_i=70 \), \( \bar{x}=10 \)
\( \sum (x_i - 10)^2=(3 - 10)^2+(8 - 10)^2+(8 - 10)^2+(11 - 10)^2+(11 - 10)^2+(13 - 10)^2+(16 - 10)^2 \)
\( =49 + 4+4+1+1+9+36=104 \)
\( s_b=\sqrt{\frac{104}{6}}\approx\sqrt{17.333}\approx4.163 \)

Data set (c) corrected:

\( \sum x_i=70 \), \( \bar{x}=10 \)
\( \sum (x_i - 10)^2=(3 - 10)^2+(6 - 10)^2+(9 - 10)^2+(10 - 10)^2+(12 - 10)^2+(14 - 10)^2+(16 - 10)^2 \)
\( =49 + 16+1+0+4+16+36=122 \)
\( s_c=\sqrt{\frac{122}{6}}\approx\sqrt{20.333}\approx4.51 \)

Now, data set (a) has \( s_a\approx3.958 \), data set (b) \( s_b\approx4.163 \), data set (c) \( s_c\approx4.51 \). So the variation (standard deviation) is not the same. The smallest variation is data set (a), and the largest is data set (c).

But according to the original data (with typos), if we calculate:

Data set (a) mean=10, s≈3.958

Data set (b) mean≈9.857, s≈3.934

Data set (c) mean≈9.857, s≈4.298

But the problem states "the three data sets have the same mean and range", so likely a typo in data set (b) and (c) to have sum 70. Assuming that, the order of standard deviation (variation) is (a) < (b) < (c). So the data set with smallest variation is (a), and largest is (c).

Answer:

The data set (a) has the smallest variation; the data set (c) has the largest variation. (Assuming the problem's statement of same mean and range is correct, with possible typos in data values to achieve same mean. The key is that the more clustered the data around the mean, the smaller the standard deviation