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PSYCHO15rus [73]
3 years ago
13

The employees of a firm that manufactures insulation are being tested for indications of asbestos in their lungs. The firm is re

quested to send three employees who have positive indications of asbestos to a medical center for further testing. If 30% of the employees have positive indications of asbestos in their lungs, find the probability that eleven employees must be tested in order to find three positives. (Round your answer to three decimal places.)
Mathematics
1 answer:
diamong [38]3 years ago
8 0

Answer:

0.070

Step-by-step explanation:

Y = number on trial

Y has a negative binomial distribution

r = 3

P = 30% = 0.3 probability of positive indication.

P(Y = 11) probability of 11 employees that must be tested to get 3 positives

Y-1Cr-1*p^r*q^(y-r)

Y-1 = 11-1 = 10

r-1 = 3 -1 = 2

10C2 x 0.3³x0.7⁸

45x0.027x0.05764801

= 0.070

This is the probability that 11 employees must be tested to get 3 positives.

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Answer:

a. The positive difference between Nelson's height and the population mean is: \\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

b. The difference found in part (a) is 1.174 standard deviations from the mean (without taking into account if the height is above or below the mean).

c. Nelson's z-score: \\ z = -1.1739 \approx -1.174 (Nelson's height is <em>below</em> the population's mean 1.174 standard deviations units).

d. Nelson's height is <em>usual</em> since \\ -2 < -1.174 < 2.

Step-by-step explanation:

The key concept to answer this question is the z-score. A <em>z-score</em> "tells us" the distance from the population's mean of a raw score in <em>standard deviation</em> units. A <em>positive value</em> for a z-score indicates that the raw score is <em>above</em> the population mean, whereas a <em>negative value</em> tells us that the raw score is <em>below</em> the population mean. The formula to obtain this <em>z-score</em> is as follows:

\\ z = \frac{x - \mu}{\sigma} [1]

Where

\\ z is the <em>z-score</em>.

\\ \mu is the <em>population mean</em>.

\\ \sigma is the <em>population standard deviation</em>.

From the question, we have that:

  • Nelson's height is 68 in. In this case, the raw score is 68 in \\ x = 68 in.
  • \\ \mu = 70.7in.
  • \\ \sigma = 2.3in.

With all this information, we are ready to answer the next questions:

a. What is the positive difference between Nelson​'s height and the​ mean?

The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):

\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

That is, <em>the positive difference is 2.7 in</em>.

b. How many standard deviations is that​ [the difference found in part​ (a)]?

To find how many <em>standard deviations</em> is that, we need to divide that difference by the <em>population standard deviation</em>. That is:

\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174

In words, the difference found in part (a) is 1.174 <em>standard deviations</em> from the mean. Notice that we are not taking into account here if the raw score, <em>x,</em> is <em>below</em> or <em>above</em> the mean.

c. Convert Nelson​'s height to a z score.

Using formula [1], we have

\\ z = \frac{x - \mu}{\sigma}

\\ z = \frac{68\;in - 70.7\;in}{2.3\;in}

\\ z = \frac{-2.7\;in}{2.3\;in}

\\ z = -1.1739 \approx -1.174

This z-score "tells us" that Nelson's height is <em>1.174 standard deviations</em> <em>below</em> the population mean (notice the negative symbol in the above result), i.e., Nelson's height is <em>below</em> the mean for heights in the club presidents of the past century 1.174 standard deviations units.

d. If we consider​ "usual" heights to be those that convert to z scores between minus2 and​ 2, is Nelson​'s height usual or​ unusual?

Carefully looking at Nelson's height, we notice that it is between those z-scores, because:

\\ -2 < z_{Nelson} < 2

\\ -2 < -1.174 < 2

Then, Nelson's height is <em>usual</em> according to that statement.  

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