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Chapter 5: Problem 34

In pharmacologic research a variety of clinical chemistry measurements are routinely monitored closely for evidence of side effects of the medication under study. Suppose typical blood-glucose levels are normally distributed, with mean \(=90 \mathrm{mg} / \mathrm{dL}\) and standard deviation \(=38 \mathrm{mg} / \mathrm{dL}\). Frequently, tests that yield abnormal results are repeated for confirmation. What is the probability that for a normal person a test will be at least 1.5 times as high as the upper limit of normal on two separate occasions?

Short Answer

Expert verified
The probability of a test result being at least 1.5 times the upper limit twice is approximately 1.42%.

Step by step solution

01

Define the normal limits

The typical mean blood-glucose level is given as 90 mg/dL with a standard deviation of 38 mg/dL. For normally distributed data, 'upper limit' generally refers to the boundary for what is considered normal, which for this example can be set using, say, the 95th percentile. However, assuming 'upper limit' intentionally excludes percentile cutoffs, we start by setting a more arbitrary 'upper limit' at the mean, 90 mg/dL.
02

Calculate the threshold for abnormal results

To find the threshold for what is 1.5 times this 'upper limit', we calculate \( 1.5 \times 90 = 135 \) mg/dL. Hence, any reading above 135 mg/dL would be considered an abnormal result that is 1.5 times the upper limit.
03

Compute the z-score

The blood-glucose reading we want to consider is 135 mg/dL. To find out how frequently this occurs in a normal distribution, compute the z-score with \( z = \frac{135 - 90}{38} \approx 1.184 \) indicating the number of standard deviations 135 mg/dL is above the mean.
04

Find the probability of one abnormal result

Using the z-table, find the probability for \( z = 1.184 \). The cumulative probability for \( z = 1.184 \) is approximately 0.881. Thus, the probability of a result exceeding this z-score, an abnormal reading, is \( 1 - 0.881 = 0.119 \).
05

Compute the probability for two independent abnormal results

Since the results on two different occasions are independent, the probability of both tests being abnormal is the product of the individual probabilities: \( 0.119 \times 0.119 = 0.0142 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Normal Distribution in Biostatistics
In biostatistics, the concept of normal distribution is central to understanding data related to clinical research, such as blood-glucose levels. Normal distribution, also known as Gaussian distribution, is a bell-shaped curve that shows how data is distributed around the mean. Most values cluster around a central peak and are symmetrically distributed from the mean. This distribution helps researchers understand the probability of deviations from the norm.

In practical terms, this means that in a clinical setting, such as measuring blood-glucose levels, most readings will be close to the average value, with fewer readings in the tails of the distribution. This curve is defined by two parameters: the mean (average) and the standard deviation (a measure of how spread out the numbers are). These parameters are crucial for identifying what's considered 'normal' or 'abnormal' in patient results. For blood-glucose levels with a mean of 90 mg/dL and a standard deviation of 38 mg/dL, these figures help determine thresholds for further investigation into potential side effects of medications.
Introduction to Z-Score Calculation
To interpret clinical data accurately, such as identifying abnormal blood-glucose levels, one can use the z-score calculation. The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It tells you how many standard deviations away a particular score is from the mean.

In our scenario, blood-glucose levels are normally distributed with a mean of 90 mg/dL. By using the z-score formula:
  • Calculate the z-score with the formula: \[ z = \frac{X - \mu}{\sigma} \]where:
    • \( X \) is the blood-glucose level of interest (for instance, 135 mg/dL),
    • \( \mu \) is the mean (90 mg/dL), and
    • \( \sigma \) is the standard deviation (38 mg/dL).
This helps in calculating how unusual (or not) a 135 mg/dL reading is. By calculating the z-score for 135 mg/dL, we find it approximately equals 1.184, indicating it is 1.184 standard deviations above the mean.
The Role of Clinical Chemistry in Healthcare
Clinical chemistry is a critical field within laboratory medicine that involves monitoring biochemical and molecular changes in patients. This helps in identifying and evaluating diseases and conditions, such as diabetes, indicated through blood-glucose tests. Without clinical chemistry, it would be challenging to establish and monitor treatment efficacy or medication side effects.

For instance, in pharmacologic research, clinical chemistry tests such as blood glucose measurements are vital. They can detect side effects or abnormal reactions to medications early. When abnormalities are suspected, tests might be repeated to confirm the results. For researchers, understanding the likelihood of readings being true abnormalities versus random distribution errors allows for informed decision-making in patient care.

In summary, clinical chemistry not only aids in routine diagnostics but also plays a pivotal role in all medical disciplines for better patient outcomes by ensuring that test results are both reliable and relevant to patient treatment plans.

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Most popular questions from this chapter

A study was conducted assessing the effect of observer training on measurement of blood pressure based on NHANES data from \(1999-2000\) (Ostchega, et al., [1] ). A goal was that the difference in recorded blood pressure between the observer and trainer be \(\leq 2 \mathrm{mm}\) Hg in absolute value. It was reported that the mean difference in systolic blood pressure (SBP) between observers and trainers (i.e., mean observer SBP minus trainer SBP) = mean (\Delta) was \(0.189 \mathrm{mm} \mathrm{Hg}\) with \(\mathrm{sd}=2.428 \mathrm{mm} \mathrm{Hg}\). If we assume that the distribution of \(\Delta\) is normally distributed, then what is the 90 percentile (i.e., the upper decile) and 10 percentile (i.e., the lower decile) of the distribution?

Well-known racial differences in blood pressure exist between Caucasian and African American adults. These differences generally do not exist between Caucasian and African American children. Because aldosterone levels have been related to blood-pressure levels in adults in previous research, an investigation was performed to look at aldosterone levels among African American children and Caucasian children [2]. If the mean plasma-aldosterone level in African American children was 230 pmol/L with a standard deviation of \(203 \mathrm{pmol} / \mathrm{L},\) then what percentage of African American children have levels \(\leq 300\) pmol/L if normality is assumed?

Retinitis pigmentosa (RP) is a genetic ocular disease that results in substantial visual loss and in many cases leads to blindness. One measure commonly used to assess the visual function of these patients is the Humphrey \(30-2\) visualfield total point score. The score is a measure of central vision and is computed as a sum of visual sensitivities over 76 locations, with a higher score indicating better central vision. Normals have an average total point score of 2500 db (decibels), and the average 37 -year-old RP patient has a total point score of 900 db. A total point score of \(<250\) db is often associated with legal blindness. Longitudinal studies have indicated that the change in total point score over N years of the average RP patient is normally distributed with mean change \(=45 \mathrm{N}\) and variance of change \(=1225 \mathrm{N}\) (Assume the total point score is measured without error; hence, no continuity correction is needed.) What is the probability that a patient will change by \(\geq 200\) db over 5 years?

Well-known racial differences in blood pressure exist between Caucasian and African American adults. These differences generally do not exist between Caucasian and African American children. Because aldosterone levels have been related to blood-pressure levels in adults in previous research, an investigation was performed to look at aldosterone levels among African American children and Caucasian children [2]. If the mean plasma-aldosterone level in Caucasian children is 400 pmol/L with standard deviation of 218 pmol/L, then what percentage of Caucasian children have levels \(\leq\) \(300 \mathrm{pmol} / \mathrm{L}\) if normality is assumed?

A study was recently published in Western Australia on the relationship between method of conception and prevalence of major birth defects (Hansen et al. [8]). The prevalence of at least one major birth defect among infants conceived naturally was \(4.2 \%,\) based on a large sample of infants. Among 837 infants bom as a result of invitro fertilization (IVF), 75 had at least one major birth defect. Does an unusual number of infants have at least one birth defect in the IVF group? Why or why not? (Hint: Use an approximation to the binomial distribution.)
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