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olya-2409 [2.1K]
3 years ago
5

The animal department wants to estimate baboon population. So they paint 50 baboons with a mark. These baboons were then release

d into the jungle. After two months 300 baboons were caught. Among these baboons 25 were marked. To the nearest whole number what is the best estimate for the baboon population?
Mathematics
1 answer:
Svet_ta [14]3 years ago
3 0

Answer:

  600 baboons

Step-by-step explanation:

Half the number of marked baboons showed up in the sample, so we expect the sample size to be about half of the total population. 300 is half of 600.

___

Many assumptions are involved in such an experiment and analysis. Not the least of which are the assumptions that the marked baboons end up evenly distributed through the population, and that the sample is a fair sample of the entire population.

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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      

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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} }

=>      \sigma  =   0.02106

Generally the  probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

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

  Here  \^ p is the sample proportion  of persons with a college degree.

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

  P(Z <  2.37 ) = 0.9911

and

  P(Z <  - 2.37 ) = 0.0089

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=>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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Step-by-step explanation:

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