Understanding the Problem

Initially, there are 23 passengers with an average (mean) weight of 64 kg and a standard deviation of 10 kg. At a stop, 5 passengers get off; no new passengers board, so now there are 18 passengers on the bus.

The question asks: what happens to the standard deviation of the weights of the remaining passengers?

Standard deviation measures how spread out the data (weights) are around the mean. It depends on:

• The individual values (weights of each passenger).
• The mean of the remaining passengers.
• How far each remaining weight is from that new mean.

Since we are only told the original mean and standard deviation, but not the actual weights of the 5 passengers who got down, we cannot compute or predict the new standard deviation with certainty.

Why Each Option Is (In)correct

a. It will decrease

This would be true only if the 5 passengers who got down were the heaviest and/or lightest ones (outliers far from the mean). Outliers increase data spread, so removing them tends to reduce the standard deviation. But since the question doesn’t specify this, we cannot conclude that SD will decrease.

b. It will not change

This would mean that the spread remains exactly the same after removing 5 people. Usually, removing any data point changes the mean and deviations, so standard deviation almost always changes unless by coincidence. Hence, it’s unlikely and indeterminable here.

c. It will increase

This would happen only if the 5 passengers who left were close to the original mean. Removing values near the mean can make the remaining data more spread out relative to the new mean, increasing the standard deviation — but we do not have that information.

d. The answer cannot be determined from the given information

This is the correct choice. The change in standard deviation depends on which 5 passengers left:

• If they were extreme (very heavy/light), SD would likely decrease.
• If they were close to the mean, SD might increase.
• It could also stay roughly the same by coincidence.

Since the problem gives only overall mean and SD, without individual weights, the answer cannot be determined.

What Is Standard Deviation?

Standard deviation (SD) measures how spread out data points are around the mean. A higher SD means greater scatter, while a lower SD means the data are close to the average.

SD = √( Σ(xᵢ – x̄)² / (n – 1) )

Where:

• xᵢ = weight of each passenger
• x̄ = mean weight
• n = number of passengers

When 5 passengers leave, both n and the data values change, so SD can go up, down, or stay the same depending on which passengers were removed.

How Removing Data Points Affects Standard Deviation

• If outliers are removed: The remaining data are closer to the mean → SD decreases.
• If values near the mean are removed: The remaining data become more spread out → SD increases.
• If a mix is removed: The final SD depends on the exact values → outcome unpredictable.

So, without knowing which passengers got down, the direction of change in SD cannot be established.

Why the Answer Is “Cannot Be Determined”

In this bus problem:

• Initial: 23 passengers, mean = 64 kg, SD = 10 kg.
• After: 18 passengers, but we don’t know which 5 left or their weights.

Because we lack information about whether the 5 who left were heavy, light, or average, we cannot compute the new sum of squared deviations from the mean. Thus, the new standard deviation cannot be determined.

Common Misconceptions

• “Removing people always reduces spread”: False; removing average-weight passengers can increase spread.
• “SD depends only on sample size”: SD depends on actual data values, not just count.
• “The ±10 kg tells us enough”: The original SD doesn’t reveal who left, so it’s insufficient.

Conclusion

When 5 passengers get down from a bus of 23 with mean 64±10 kg:

• SD could decrease if outliers left.
• SD could increase if average-weight passengers left.
• SD could stay roughly the same by coincidence.

Since we do not know which passengers left, the correct answer is: the answer cannot be determined from the given information.