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galben [10]
3 years ago
10

Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,

69, 71, 78, 80, 81, 82, 82
Mathematics
2 answers:
Firdavs [7]3 years ago
8 0

C. or 8, 17, 18, 20, 20, 21, 23, 26, 31, 39

umka21 [38]3 years ago
3 0

It's hard to tell where one set ends and the next starts.  I think it's

A. 25, 36, 44, 51, 62, 77

B. 3, 3, 3, 7, 9, 9, 10, 14

C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39

D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.

A. 25, 36, 44, 51, 62, 77

That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.

B. 3, 3, 3, 7, 9, 9, 10, 14

Average around 7, sigma around 4, within 2 sigma, seems ok.

C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39

Average around 20, sigma around 8, that 39 is hanging out there past two sigma.  Let's reserve judgement and compare to the next one.


D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82

Average around 74, sigma 8, seems very tight.

I guess we conclude C has the outlier 39.  That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.  



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3 years ago
A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic
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Answer:

(a) The proportion of women who are tested, get a negative test result is 0.82.

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Step-by-step explanation:

The Bayes' theorem states that the conditional probability of an event <em>E</em>_{i}, of the sample space <em>S,</em> given that another event <em>A</em> has already occurred is:

P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\liits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}

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P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}

Denote the events as follows:

<em>X</em> = fetus have a chromosome abnormality.

<em>Y</em> = the test is positive

The information provided is:

P(X)=0.04\\P(Y|X)=0.90\\P(Y^{c}|X^{c})=0.85

Using the above the probabilities compute the remaining values as follows:

P(X^{c})=1-P(X)=1-0.04=0.96

P(Y^{c}|X)=1-P(Y|X)=1-0.90=0.10

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(a)

Compute the probability of women who are tested negative as follows:

Use the law of total probability:

P(Y^{c})=P(Y^{c}|X)P(X)+P(Y^{c}|X^{c})P(X^{c})

          =(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82

Thus, the proportion of women who are tested, get a negative test result is 0.82.

(b)

Compute the value of P (X|Y) as follows:

Use the Bayes' theorem:

P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}

             =\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}

             =0.20

Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

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