Let x be an observation from a Normal Distribution with mean µ and standard deviation σ. The standardized score (z-score) is defined byz = (x – µ)/σFor the following values of x, µ and σ, find z. Give your answer correct to 2 decimal places.x = 112 µ = 100 σ = 12.5

Question

Let x be an observation from a Normal Distribution with mean µ and standard deviation σ. The standardized score (z-score) is defined byz = (x – µ)/σFor the following values of x, µ and σ, find z. Give your answer correct to 2 decimal places.x = 112 µ = 100 σ = 12.5

BlackTom AI · Accepted Answer

To begin, I will restate the problem in my own words and identify what is being asked: compute the z-score z = (x − μ)/σ for the given values x = 112, μ = 100, σ = 12.5, and report to two decimal places.
Option a: z = 104. This would imply a z-score of 104, which is far from the standard-normal scaling where z is a standardized deviation in units of standard deviation. It seems to confuse the raw value x with the standardized score, so this choice is not consistent with the formula.
Option b: z = 106. Similar to option a, 106 would place z far outside the typical range and does not align with subtracting the mean and dividing by the standard deviation. It appears to be another miscalculation that treats the result as a scaled raw value rather than a standardized one.
Option c: Other. This would be chosen if none of the listed numerical options matched the correct computation. Since we can perform the arithmetic, we should verify whether the actual computed z matches any specific listed form before deciding this is appropriate.
Option d: z = 0.96. This value looks like a properly computed z-score when doing (112 − 100) / 12.5, but we need to confirm the arithmetic to ensure two-decimal accuracy. If the division yields 0.96, this would be a correct two-decimal representation.
Option e: z = 0. This would occur only if x equaled μ, i.e., 112 = 100, which is not true here. Therefore this option cannot be correct given the numbers provided.
Now, performing the actual calculation: subtract the mean from x: 112 − 100 = 12. Then divide by the standard deviation: 12 / 12.5 = 0.96. The result is a positive z-score less than 1, indicating the observation lies slightly above the mean in units of σ. Expressing to two decimal places yields 0.96, which matches the form shown in one of the options.
In light of the above, the option that aligns with the computed z-score to two decimals is the one that states z = 0.96, while the others fail for different reasons (either incorrect arithmetic, misunderstanding of z-score meaning, or inconsistency with the two-decimal formatting).

My LMS Subjects Maths for Biomedicine DECIMALS QUIZ

Let x be an observation from a Normal Distribution with mean µ and standard deviation σ. The standardized score (z-score) is defined byz = (x – µ)/σFor the following values of x, µ and σ, find z. Give your answer correct to 2 decimal places.x = 112 µ = 100 σ = 12.5

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