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tankabanditka [31]
3 years ago
10

Pls only answer if you know it:) The numbers are 19, 36, 24, 40, 17, 43, 30, 31, 37, 28, 45, 32, 16. What are the first quartile

, median and third quartile to this data set?
a) 21.5, 31, 37
b) 21.5, 31, 38.5
c) 24, 31, 38.5
d) 24, 31, 37
Mathematics
2 answers:
charle [14.2K]3 years ago
6 0

Answer:

The answer is B

Step-by-step explanation:

You order them from least to greatest- 16, 17, 19, 24, 28, 30, 31, 32, 36, 37, 40, 43, 45.

Then you find the middle number or the median- 31

Then you go to the left of the medain and get the first quartile- 19, 24 (You have to find the mean of these two numbers- 21.5)

Then you go to the right of the medain and get the third quartile- 37, 40 (Again you must find the mean of these two numbers- 38.5)

Last you would put them in the order needed.

Hope this helps! :)  

NNADVOKAT [17]3 years ago
5 0

Answer:

The answer is b

Step-by-step explanation:

The reason being that I just did it myself on a lesson and got it right.

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Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that
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Answer:

The value is  P(| \^ p -  p| < 0.05 ) = 0.9822

Step-by-step explanation:

From the question we are told that

    The population proportion is  p =  0.52

     The sample size is  n  =  563      

Generally the population mean of the sampling distribution is mathematically  represented as

           \mu_{x} =  p =  0.52

Generally the standard deviation of the sampling distribution is mathematically  evaluated as

       \sigma  =  \sqrt{\frac{ p(1- p)}{n} }

=>      \sigma  =  \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }

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            P(| \^ p -  p| < 0.05 ) =  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 ))

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So

 P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma }  < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )

Here  

    \frac{[\^p - p] - p}{\sigma }  = Z (The\ standardized \  value \  of\  (\^ p - p))

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=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[ -2.37 <  Z  < 2.37 ]

=>  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(Z <  2.37 ) - P(Z < -2.37 )

From the z-table  the probability of  (Z <  2.37 ) and  (Z < -2.37 ) is

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and

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So

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) =0.9911-0.0089

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = 0.9822

=> P(| \^ p -  p| < 0.05 ) = 0.9822

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