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Suppose a student completes an experiment with an average value of 2.9 mL and a calculated standard deviation of 0.71 mL. What is the minimum value within a 1 SD range of the average

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Answer:

The correct answer is 2.2 mL.

Explanation:

Given:

Average: 2.9 mL

SD: 0.71 mL

We can define a 1 SD range in which the value of volume (in mL) will be comprised:

Volume (mL) = Average ± SD = (2.9 ± 0.7) mL

Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL

Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL

Thus, the minimum value within a 1 SD range of the average is 2.2 mL

The minimum value within 1 SD is 2.19 mL

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x\ is\ raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]

Given that μ = 2.9 mL, σ = 0.71 mL; hence:

The minimum value within 1 SD range = μ ± σ = 2.9 ± 0.71 = (2.19, 3.61)

Therefore the minimum value within 1 SD is 2.19 mL

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