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Olin [163]
3 years ago
10

Explain why the measure was selected to report on the data set. The mean price of homes sold in the area is $197,000. Home Price

s $124,000 $99,000 $130,000 $122,000 $450,000 $106,000 $114,000 $497,000 $131,000 The mean price make homes seem to have a higher value than the median price because of the outliers. The mean price make homes seem to have a lower value than the median price because of the outliers. The mean and the median of the data set have the same value. The mean and the median of the data can not be calculated.
Mathematics
1 answer:
aniked [119]3 years ago
7 0

Answer:

Only one says the mean s higher than the median, so choose that one.  

Step-by-step explanation:

Putting them in order gets yout he following list

$99,000

$106,000

$122,000

$124,000

$130,000

$131,000

$114,000

$497,000

$450,000

The average changes if you increase or decrease any of the prices, thought he median does not unless you mess witht he middle value 130,000.  Since it mentions the outliers lets look at those.  The outliers are the last two because they are more than 3 times the next highest.  So, increasing or decreasing these will effet the average in the same way.  Also, keep in mind what the median is, and if the verage as is is higher or lower than it., that pretty much answers your question.

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Rule for 1 is the rule for 1 is 14 times 3.5^x and it is exponential

The second one is is also exponential and the rule is 32 times .25^x and the last one is is linear and the rule is 5x+-4
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Suppose that P(A) = 0.25 and P(B) = 0.40 . If P(A|B)=0.20 , what is P(B|A) ?
Ket [755]

Answer:

  • 0.32

Step-by-step explanation:

<u>Use of formula:</u>

  • P(A and B) = P(A)*P(B|A) and
  • P(A and B) = P(B)*P(A|B)

<u>According to above and based on given:</u>

  • P(A)P(B|A) = P(B)P(A|B)
  • P(B|A) = P(A|B)*P(B)/P(A)
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8 0
3 years ago
There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=
yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = \frac{12P4}{12^{4} }

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = \frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }

P(X=2) = 0.401

P(X=3) = \frac{4C3 * 12P2}{12^{4} }

P(x=3) = 0.025

P(X=4) = \frac{12}{12^{4} }

P(X=4) = 0.001

Mean, \mu = \sum xP(x)

\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879

Standard deviation, SD = \sqrt{\sum x^{2} P(x) - \mu^{2}}  \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2}  * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}

SD = 1.033

4 0
3 years ago
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