91Ó°ÊÓ

Blood plasma proteins of patients with malignant breast tumors differ from proteins of healthy people in their solubility in the presence of various polymers. When the polymers dextran and poly(ethylene glycol) are mixed with water, a two-phase mixture is formed. When plasma proteins of tumor patients are added, the distribution of proteins between the two phases is different from that of plasma proteins of a healthy person. The distribution coefficient ( K) for any substance is defined as K =[concentration of the substance in phase[concentration of the substance in phase B ]. Proteins of healthy people have a mean distribution coefficient of 0.75 with a standard deviation of 0.07. For the proteins of people with cancer, the mean is 0.92 with a standard deviation of 0.11.

(a) Suppose that Kwere used as a diagnostic tool and that a positive indication of cancer is taken as$\mathit{K}\mathbf{â‰\yen }\mathbf{0}\mathbf{.}\mathbf{92}$. What fraction of people with tumors would have a false negative indication of cancer because$\mathit{K}\mathbf{â‰\yen }\mathbf{0}\mathbf{.}\mathbf{92}$?

(b) What fraction of healthy people would have a false positive indication of cancer? This number is the fraction of healthy people with$\mathit{K}\mathbf{â‰\yen }\mathbf{0}\mathbf{.}\mathbf{92}$, shown by the shaded area in the graph below. Estimate an answer with Table 4 - 1 and obtain a more exact result with the NORMDIST function in Excel.

(c) Vary the first argument of the NORMDIST function to select a distribution coefficient that would identify 75% of people with tumors. That is, 75% of patients with tumors would have K above the selected distribution coefficient. With this value of K, what fraction of healthy people would have a false positive result indicating they have a tumor?

Step by step solution

01

Definition of fraction.

A fraction is a number that denotes a portion of something larger. There are two parts to it: a numerator and a denominator.

02

Find the fraction of people with tumors would have a false negative indication of cancer because K≥0.92 .

(a)

The distribution coefficient, K is utilised as a diagnostic tool, and$Kâ‰\yen 0.92$ is considered a positive indicator of cancer.

Half of persons with cancer have $Kâ‰\yen 0.92$and a false negative indication of cancer.

03

Find the fraction of healthy people would have a false positive indication of cancer.

(b)

The fraction of healthy patients with$Kâ‰\yen 0.92$ would receive a cancer diagnosis based on a false positive result.

To estimate an answer with Table 4 - 1 , you must convert x to a z value.

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ ≈\frac{x-\stackrel{¯}{x}}{s} \\ =\frac{0.92-0.75}{0.07} \\ =2.4 \end{gathered}$$

The area of a range determines the likelihood of measuring in that range.

Table 4 - 1 shows the area under each part of the Gaussian curve.

The area from the mean to is z = 2.4 is 0.4983 , as shown in Table 4 - 1 , ordinate and area for the normal (Gaussian) error curve.

Above z = 2.4% , the area is 0.5 - 0.4983 = 0.008 .

As a result, 0.8% of healthy patients will have a false positive cancer signal.

The table below shows the outcome of using the NORMDIST function in Excel.

04

Find the fraction of healthy people would have a false positive result indicating they have a tumor.

(c)

In cells A7 -A13 , a distribution coefficient is chosen that would indicate 75% of persons with malignancies.

The value of K yields an area of 0.25 , implying that 25% of cancer patients will go unidentified.

The number 0.84 yields a 0.2335 area, while the value 0.85 yields a 0.2623 area. However, the number K = 0.845 results in an area of 0.25 .

The area below K = 0.845 in cell B15 is 0.912632 .

1 - 0.912632 = 0.087368 is the region above K = 0.845 .

As a result, 8.7% of healthy patients would get a false positive result suggesting a malignancy.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!