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Zorluk: OrtaData Types, Distributions, and Descriptive Statistics

A cross-sectional study is conducted to evaluate the intraocular pressure (IOP) in a population of 5,0005,000 healthy adults over the age of 4040. The collected IOP measurements are determined to follow a normal distribution with a mean of 1515 mm Hg. The researchers note that approximately 800800 participants (16%16\% of the cohort) have an IOP greater than 1818 mm Hg.

Based on these findings, what is the variance of the intraocular pressure in this population?

Cevap: 9 mm Hg^2

Cevap

9
The correct answer is 99. In a normal distribution, the empirical rule dictates that approximately 68%68\% of values fall within one standard deviation (SD) of the mean. Because the distribution is symmetrical, the remaining 32%32\% of values fall outside this range, with 16%16\% located in the upper tail (above +1+1 SD) and 16%16\% in the lower tail (below −1-1 SD). The study states that 16%16\% of participants have an IOP greater than 1818 mm Hg, meaning 1818 mm Hg represents exactly +1+1 SD from the mean of 1515 mm Hg. The standard deviation is therefore 18−15=318 - 15 = 3 mm Hg. Variance is mathematically defined as the square of the standard deviation (SD2SD^2). Therefore, the variance is 32=93^2 = 9.

Adım Adım Çözüm

1
Determine the number of standard deviations from the mean for an IOP of 18 mm Hg.
18 mm Hg represents exactly +1 SD.
According to the empirical rule for normal distributions, 68% of the data falls within +/- 1 SD. This leaves 32% in the tails, meaning 16% of the data lies above +1 SD. Since 800 out of 5,000 participants (16%) have an IOP above 18 mm Hg, 18 must be the value at +1 SD.
2
Calculate the standard deviation (SD).
SD = 3 mm Hg.
The standard deviation is the absolute difference between the mean (15) and the value at +1 SD (18). 18 - 15 = 3.
3
Calculate the variance.
Variance = 9.
Variance is defined as the square of the standard deviation. 3^2 = 9.

Anahtar Kavram

Application of the empirical (68-95-99.7) rule to identify the standard deviation from population percentiles, and calculating variance as the square of the standard deviation.
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