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kvasek [131]
3 years ago
8

Eric randomly surveyed 150 adults from a certain city and asked which team in a contest they were rooting​ for, either North Hig

h School or South High School. From the results of his​ survey, Eric obtained a​ 95% confidence interval of​ (0.52,0.68) for the proportion of all adults in the city rooting for North High. What proportion of the 150 adults in the survey said they were rooting for North High​ School?
A. 0.646

B. 0.60

C. Somewhere between 0.52 and 0.68, but the exact proportion cannot be determined.

D. This information cannot be determined from the information given in the problem
Mathematics
1 answer:
prohojiy [21]3 years ago
6 0

Answer:

B. 0.60

Step-by-step explanation:

Let be

X = adults in the survey said they were rooting for North High School

µ = mean

σ = standard deviation

let be

X1 = 0.52

X2 = 0.68

We know that a 95% confidence interval correspond to z values of -1.96 and 0.96

Now we have two equations to find µ and σ

-1.96 = (0.52 –µ)/σ  ….. equation 1

1.96 = (0.68 –µ)/σ   ….. equation 2

Dividing equation 1 and equation 2

-1.0 = (0.52 –µ)/ (0.68 –µ)

-1.0 (0.52 –µ) = (0.68 –µ)  

-0.52 + µ =0.68 - µ

µ + µ = 0.68 + 0.52

2 µ = 1.2

µ = 0.6

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In order to find this, we first need to see how many pounds of rations a single soldier eats in a week. To do this, we take the total eaten in 3 weeks and divide by 3. Then we divide by the number of soldiers.


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Now we look to see how many soldiers we will have in total after adding. There are 520 to start and we add 780 to get 1300 total. Next we multiply that by the total per soldier per week.


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Then we have to multiply by the 5 weeks that battalion will be there.


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