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garri49 [273]
4 years ago
7

How Important Is Regular Exercise? In a recent poll1 of 1000 American adults, the number saying that exercise is an important pa

rt of daily life was 753. Use StatKey or other technology to find a 90% confidence interval for the proportion of American adults who think exercise is an important part of daily life. Click here to access StatKey. Round your answer to three decimal places. The 90% confidence interval is Enter your answer; the 90% confidence interval, value 1
Mathematics
1 answer:
Elza [17]4 years ago
6 0

Answer:

Step-by-step explanation:

Given that in a recent poll 1 of 1000 American adults, the number saying that exercise is an important part of daily life was 753.

Sample proportion = \frac{753}{1000} =0.753

To find 90% confidence interval we must find std error of proportion first

As per central limit theorem, sample proportion will follow a normal distribution with mean = 0.753

Std error of proportion= \sqrt{0.753*0.247/1000} \\=0.01363

90% Z critical value = 1.645

Margin of error = 1.645*0.01363 = 0.0224

Confidence interval lower bound = 0.753-0.0224=0.731

Upper bound = 0.775

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3 years ago
In a certain school there are 180 pupils in the year7,110 pupils study french,88 study German , and 65 indonesian.40 pupils stud
MAXImum [283]

Answer:

(a) All three languages  - 9

(b) Indonesian only  - 10

(c) none of the languages - 12

(d) At least one language  - 168

(e) Either one or two of the three languages - 159​

Step-by-step explanation:

The question require the knowledge of Set theory and its Formulas.

Total pupils = 180

French (Let F) = 110  

German (Let G) = 88

Indonesian (Let I) =  65

French and German (F intersection G) =  40

German and Indonesian (G intersection I) =  38

French and Indonesian (F intersection I)  = 26

German only  = 19

(a) All three languages

We are given only German speaking people are 19

Only German = n(G) - n( F intersection G) - n(G intersection I) + n(F intersection G intersection I)

 19 = 88 - 40-38 + n(G intersection I) + n(F intersection G intersection I)

n(G intersection I) + n(F intersection G intersection I) = 9

n(G intersection I) + n(F intersection G intersection I) represents the number of pupils speaking who study all the three languages.

(b)Indonesian only

   n( I) - n(G intersection I) + n(F intersection I) + n(G intersection I) + n(F intersection G intersection I)

65 - 38 - 26 + 9 = 10

So 10 pupils speak Indonesian only

(c)none of the languages

It will be equal to Total - pupils speaking any of the three languages

Any of the three languages = (F union G union I)

  = n(F) +n(G) + n(I) -n(F intersection G) - n(G intersection I) -n(F intersection I)

    +  n(F intersection G intersection I)

n(G intersection I)

  = 110 + 88 + 65 -40 -38-26 + 9

  = 263 - 95

   = 168

So 168 pupils speak any of the three languages

None speakers = 180 - 168 = 12 pupils

So 12 pupils do not speak any of the three languages.

(d)at least one language

At least one language has the meaning that the person can either speak one two or all three languages, so it will be same as we proceeded above

Any of the three languages = (F union G union I)

  = n(F) +n(G) + n(I) -n(F intersection G) - n(G intersection I) -n(F intersection I)

    +  n(F intersection G intersection I)

n(G intersection I)

  = 110 + 88 + 65 -40 -38-26 + 9

  = 263 - 95

   = 168

(e) either one or two of the three languages​

Pupils can speak one or two language but not all the three so will subtract all the three language speaker from the total.

  Total speaker = 168

 All three languages speaker  = 9

 Either one or two of the three languages​ speaker = 168 - 9 = 159

8 0
3 years ago
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