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Elena L [17]
3 years ago
8

Consumption of fast food is a topic of interest to researchers in the field of nutrition. An article reported that 1880 of those

in a random sample of 6765 U.S. children indicated that on a typical day, they ate fast food. Estimate p, the proportion of children in the United States who eat fast food on a typical day. (Round your answer to four decimal places.)
Mathematics
1 answer:
frez [133]3 years ago
3 0

Answer:

the proportion is 0.2779, that is a 27.79%

Step-by-step explanation:

The article reported that 1880 children ate fast food from a sample of 6765 children. We can use the rule of three to find the proportion of children that ate fast food

  • from 6765 we have 1880 children that ate fast food
  • from 100 children we have 1880 * (\frac{100}{6765}) = 27.79 children that ate fast food

that gives us that a 27.79% of the children ate fast food on a typical day. We can obtain the proportion by dividing the percentage by 100, that gives us a proportion of 27.79/100 = 0.2779.

I hope this helped you!

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(i) 0.15708

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(iii) 3

Step-by-step explanation:

Given that, 99% of people who fracture or dislocate a bone see a doctor for that condition.

There is only two chance either the person having fracture or dislocation of bone will either see the doctor or not.

As per previous data, if one person got a fracture or dislocation of bone, the chance of seeing the doctor is 0.99. Assuming this chance is the same for every individual, so the total number of people having fractured or dislocated a bone can be considered as Bernoulli's population.

Let p be the probability of success represented by the chances of not seeing a doctor by any one individual having fractured or dislocated a bone.

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(ii) The probability that fewer than four of them did not see a doctor

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=\binom{400}{0}(0.01)^0(0.99)^{400}+\binom{400}{1}(0.01)^1(0.99)^{399}+\binom{400}{2}(0.01)^2(0.99)^{398}+\binom{400}{3}(0.01)^3(0.99)^{397}

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(iii) The expected number of people who would not see a doctor

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=300\times 0.01

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