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Selecting the Correct Normal Curve Graph Based on Mean and Standard Deviation

Why was (A) selected in the end for this problem?

Option (A) was selected because it is the only graph that satisfies both the positional and shape conditions derived from the means and standard deviations. The mean of School A(60)A (60) is less than School B(65)B (65), placing A's curve to the left. The standard deviation of School A(10)A (10) is greater than School B(5)B (5), making A's curve wider and shorter, while B's curve is narrower and taller. Option (A) correctly depicts the thick line (School A) as wider and shorter on the left, and the thin line (School B) as narrower and taller on the right.

Conditions

  • The thick line represents School A and the thin line represents School B.
  • School A has mean 60 and standard deviation 10.
  • School B has mean 65 and standard deviation 5.

Reasoning, step by step

  1. Compare the means: 60<6560 < 65, so School A's curve must be to the left of School B's curve.
  2. Compare the standard deviations: 10>510 > 5, so School A's curve must be wider and shorter than School B's curve.
  3. Evaluate the options based on these two criteria.
  4. Notice that option (B) has curves of equal height, which contradicts the different standard deviations.
  5. Confirm that option (A) shows the thick line (School A) wider and shorter on the left, and the thin line (School B) narrower and taller on the right.
  6. Select option (A) as the correct answer.

Example

The video states: 'Combine the checks: School A’s 60 and 10 match the broad, low thick curve on the left; School B’s 65 and 5 the narrow, high thin curve on the right. Diagram A fits.'

Common misconceptions

  • Choosing an option based solely on the horizontal position without checking the curve shapes.
  • Assuming that the peak heights can be equal for different standard deviations.

Watch the explanation

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.