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dmitriy555 [2]
3 years ago
12

BRAINLIEST look at the picture for the question, also please leave a small explanation :)

Mathematics
1 answer:
saul85 [17]3 years ago
5 0

Answer:

C

Step-by-step explanation:

the explanation is there :)

hope this helps <3

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(a) Margin of error ( E) = $2,000 , n = 54

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(c)   Margin of error ( E) = $500 , n= 864

Step-by-step explanation:

Given -

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Z_{\frac{\alpha}{2}} =  Z_{\frac{.05}{2}} = 1.96

let sample size is n

(a) Margin of error ( E) = $2,000

Margin of error ( E)  = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

                           E   = Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

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n =  54.0225

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(b)   Margin of error ( E) = $1,000

          E     = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

         1000   =  Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}

n = 216

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  500 = Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}

n = 864

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