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Anuta_ua [19.1K]
4 years ago
15

The housing market has recovered slowly from the economic crisis of 2008.​ Recently, in one large​ community, realtors randomly

sampled 38 bids from potential buyers to estimate the average loss in home value. The sample showed the average loss was ​$9379 with a standard deviation of ​$3000. Suppose a​ 95% confidence interval to estimate the average loss in home value is found.
​a) Suppose the standard deviation of the losses had been ​$9000 instead of ​$3000.
b) What would the larger standard deviation do to the width of the confidence interval​ (assuming the same level of​ confidence)?
Mathematics
1 answer:
Gwar [14]4 years ago
3 0

Answer:

Step-by-step explanation:

Given that the housing market has recovered slowly from the economic crisis of 2008.​ Recently, in one large​ community, realtors randomly sampled 38 bids from potential buyers to estimate the average loss in home value.

s = sample std deviation = 3000

Sample mean = 9379

Sample size n = 38

df = 37

Std error of sample mean = \frac{s}{\sqrt{n} } \\=486.66

confidence interval 95% = Mean ± t critical * std error

=Mean ±1.687*486.66 = Mean ±821.003

=(8557.997, 10200.003)

a) If std deviation changes to 9000 instead of 3000, margin of error becomes 3 times

Hence 2463.008

b) The more the std deviation the more the width of confidence interval.

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Answer:

p ( X = 1 ) = 0.6 , p ( X = 2 ) = 0.3 , p ( X = 3 ) = 0.1

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Step-by-step explanation:

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                       Color                  Number of balls

                       White                            3

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2) A red ball is drawn on the first draw and a white ball is drawn on the second trial ( X = 2 ). The probability of drawing a red ball first would be:

      p ( Red on first trial ) = ( Number of red balls ) / ( Total number of balls )

      p ( Red on first trial ) = ( 2 ) / ( 5 )

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   p ( White on second trial ) = ( Number of white balls ) / ( number of balls left )

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                p ( X = 2 ) =  p ( Red on first trial ) *  p ( White on second trial )

                p ( X = 2 ) =  ( 2 / 5 ) * ( 3 / 4 )

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3) A red ball is drawn on the first draw and second draw and then a white ball is drawn on the third trial ( X = 3 ). The probability of drawing a red ball first would be ( 2 / 5 ). Then we are left with 4 balls in the urn, we again draw a red ball:

   p ( Red on second trial ) = ( Number of red balls ) / ( number of balls left )

   p ( Red on second trial ) = ( 1 ) / ( 4 )    

 

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   p ( White on 3rd trial ) = ( Number of white balls ) / ( number of balls left )

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                p ( X = 3 ) =  ( 2 / 5 ) * ( 1 / 4 ) * 1

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               E ( X ) = ( 1 ) * ( 0.6 ) + ( 2 ) * ( 0.3 ) + ( 3 ) * ( 0.1 )

               E ( X ) = 0.6 + 0.6 + 0.3

               E ( X ) = 1.5 trials until first white ball is drawn.

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