During a quality control inspection, five manufactured items were measured and found to ha… — GMAT

During a quality control inspection, five manufactured items were measured and found to have lengths of $44\text{ mm}$ , $48\text{ mm}$ , $50\text{ mm}$ , $52\text{ mm}$ , and $56\text{ mm}$ . What is the standard deviation, in millimeters, of the lengths of these five items?

Cevap: 4 millimeters

Cevap

The standard deviation of the lengths of the five items is 4 millimeters.

The arithmetic mean of the five measurements is 50 mm. The sum of the squared deviations from 50 is 36 + 4 + 0 + 4 + 36 = 80. Dividing 80 by 5 yields a variance of 16. Taking the principal square root of 16 gives a standard deviation of 4 mm.

Adım Adım Çözüm

Calculate the arithmetic mean of the dataset

Mean = 50

The standard deviation measures dispersion relative to the mean.

Find the squared difference of each data point from the mean

Squared deviations are 36, 4, 0, 4, and 36

Squaring ensures all deviations are non-negative and penalizes larger deviations.

Compute the mean of the squared deviations (variance)

Variance = 80 / 5 = 16

Variance is the average squared distance from the mean.

Take the non-negative square root of the variance

Standard deviation = sqrt(16) = 4

Standard deviation converts variance back to the original unit of measurement.

Anahtar Kavram

Standard Deviation Calculation for a Data Set

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