13) without calculating, match the sets of values below, one from column a and one from column b, that have the same standard deviations.
column a
a: 1,2,3,4,5
b: 2,4,6,8,10
c: 2,2,2,2,2
d: 2,6,6,6,10
e: 2,2,6,10,10
column b
f: 10,10,10,10,10,10
g: 4,6,8,10,12
h: 16,16,20,24,24
i: 16,16,20,24,24
j: 4,8,8,8,12
14) once school buses reach a certain mileage, they should be replaced. the odometer readings of a sample of buses from two different school districts are shown below. the miles are given in thousands. which argument seems the most reasonable based on the data? explain your reasoning.
district a: 99, 87, 98, 65
district b: 43, 36, 35, 250
options:
a. district a needs new buses more than district b because the mean mileage of its buses is greater.
b. district a needs new buses more than district b even though the mean mileage of its buses is lower, because the smaller standard deviation in the data suggests that more of its buses will need to be replaced sooner than those in district b.
c. district b needs new buses more than district a even though the mean mileage of its buses is lower, because the smaller standard deviation in the data suggests that more of its buses will need to be replaced sooner than those in district a.
d. district b needs new buses more than district a because the mean mileage of its buses is greater.
answer: ______
explanation: ______

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Accepted Answer

### Question 14 Analysis
To determine which district needs new buses more, we analyze the data:

- **Mean Mileage**: First, calculate the mean for each district. For District A, the odometer readings are 99, 87, 98, 65. The sum is $99 + 87 + 98 + 65 = 349$, and the mean is $\frac{349}{4} = 87.25$ thousand miles. For District B, the readings are 43, 36, 35, 250. The sum is $43 + 36 + 35 + 250 = 364$, and the mean is $\frac{364}{4} = 91$ thousand miles. Wait, but there's an outlier (250) in District B's data.
- **Standard Deviation Insight**: Standard deviation measures data spread. District B has a value (250) that's much higher than the others (43, 36, 35), so its data is more spread out (higher standard deviation). However, most of District B's buses have low mileage (35, 36, 43), while District A's buses have more consistent, relatively high mileage (65, 87, 98, 99). Wait, no—wait, the problem is about when buses reach a certain mileage, they need replacement. So lower mileage means closer to replacement? Wait, no: higher mileage means more used, so when they reach the threshold, they need replacement. Wait, the odometer readings are the mileage; when they reach the certain mileage (threshold), they are replaced. So buses with higher mileage are closer to replacement? Wait, no—if the threshold is, say, 100 thousand miles, then a bus with 99 is close, 65 is far. But District B has a bus with 250, which is way above, but others are low. District A's buses are 65, 87, 98, 99—so most are close to 100. Wait, maybe I got it wrong. Let's re-express:

Wait, the question is: "Once school buses reach a certain mileage, they should be replaced. The odometer readings of a sample of buses from two different school districts are shown below. Which argument seems the most reasonable based on the data? Explain your reasoning."

Looking at the options:

- **Option C**: "District B needs new buses more than District A even though the mean mileage of its buses is lower, because the smaller standard deviation in the data suggests that more of its buses will need to be replaced sooner than those in District A." Wait, no—wait, District B's data: 43, 36, 35, 250. The mean is 91, but the standard deviation: the values 35, 36, 43 are close, but 250 is an outlier. Wait, no—maybe the threshold is a lower number? Wait, no, the odometer readings are in thousands. Wait, maybe the threshold is, say, 50 thousand miles? Then District B's buses: 35, 36, 43 are below or close, and 250 is above. District A's: 65, 87, 98, 99 are above 50. But that doesn't make sense. Wait, maybe the key is that District A's buses have higher mileage (closer to the replacement threshold) on average, but District B has an outlier, but most of its buses have low mileage? No, the options:

Wait, let's re-express the options:

- **A**: District A needs new buses more than District B because the mean mileage of its buses is greater. (Mean of A: (99+87+98+65)/4 = 349/4 = 87.25; Mean of B: (43+36+35+250)/4 = 364/4 = 91. So A's mean is lower than B's. So A is wrong.)

- **B**: District A needs new buses more than District B even though the mean mileage of its buses is lower, because the smaller standard deviation in the data suggests that more of its buses will need to be replaced sooner than those in District B. (Wait, standard deviation of A: let's calculate variance. Deviations from mean (87.25): 99-87.25=11.75, 87-87.25=-0.25, 98-87.25=10.75, 65-87.25=-22.25. Squared deviations: 138.06, 0.06, 115.56, 495.06. Sum: 138.06+0.06+115.56+495.06=748.74. Variance: 748.74/4=187.185. Std dev: ~13.68. For B: mean 91. Deviations: 43-91=-48, 36-91=-55, 35-91=-56, 250-91=159. Squared deviations: 2304, 3025, 3136, 25281. Sum: 2304+3025=5329+3136=8465+25281=33746. Variance: 33746/4=8436.5. Std dev: ~91.85. So District A has smaller standard deviation. So if the threshold is, say, 100, then District A's buses are 65,87,98,99—most are close to 100, so more need replacement. District B has 35,36,43 (far from 100) and 250 (way above). So even though A's mean is lower (87.25 vs 91), its smaller standard deviation means the data is more clustered around the mean, so more buses are close to the replacement threshold. So Option B says: "District A needs new buses more than District B even though the mean mileage of its buses is lower, because the smaller standard deviation in the data suggests that more of its buses will need to be replaced sooner than those in District B." Wait, but earlier calculation showed A's mean is 87.25, B's is 91—so A's mean is lower. And A's standard deviation is smaller (more clustered). So if the replacement threshold is, say, 100, then A's buses (65,87,98,99) have three buses above 87, two above 90, etc., while B's have 35,36,43 (way below) and 250 (way above). So more of A's buses are close to 100, so need replacement sooner. So Option B is correct? Wait, but let's check the options again.

Wait, the options:

- **A**: Mean of A is greater? No, A's mean is 87.25, B's is 91. So A is wrong.

- **B**: District A needs more than B, even though mean is lower, because smaller std dev (more clustered) means more buses need replacement sooner. This matches.

- **C**: District B needs more than A, even though mean is lower, because smaller std dev. But B's std dev is higher, so C is wrong.

- **D**: District B needs more than A because mean is greater. B's mean is greater, but its buses are mostly low (35,36,43) except the outlier. So D is wrong.

So the correct answer is **B**.

### Final Answer (Question 14)
# Brief Explanations:
To determine which district needs new buses more, we analyze mean mileage and standard deviation:
- **Mean Mileage**: District A’s mean ($\frac{99+87+98+65}{4} = 87.25$) is lower than District B’s ($\frac{43+36+35+250}{4} = 91$).
- **Standard Deviation**: District A’s data (65, 87, 98, 99) is more clustered (smaller standard deviation) around its mean, meaning more buses are close to the replacement mileage threshold. District B’s data has an outlier (250) and most buses (35, 36, 43) are far from the threshold. Thus, even with a lower mean, District A’s buses are more likely to need replacement sooner.

# Answer:
B. District A needs new buses more than District B even though the mean mileage of its buses is lower, because the smaller standard deviation in the data suggests that more of its buses will need to be replaced sooner than those in District B.

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