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Nataly [62]
2 years ago
10

Hospital floors are usually covered by bare tiles. Carpets would cut down on noise but might be more likely to harbor germs. To

study this possibility, investigators randomly assigned 8 of 16 available hospital rooms to have carpet installed. The others were left bare. Later, air from each room was pumped over a dish of agar (a gelatinous substance obtained from various kinds of red seaweed and used in biological culture media). The dish was incubated for a fixed period, and the number of bacteria colonies were counted. The response variable in this experiment is
Mathematics
1 answer:
Keith_Richards [23]2 years ago
3 0

Answer:

Given the details of the study carried

out on Hospitals to study the possibility that carpers might be more likely to harbor

germs, in which air from rooms with carpet and those without carpet was pumped over a

dish of agar, and the number of bacteria colonies were counted after incubating

the dish for a fixed period of time. The response in this study is number of bacteria colonies in a dish.

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6 0
3 years ago
A new sample of 225 employed adults is chosen. Find the probability that less than 7.1% of the individuals in this sample hold m
arsen [322]

Answer:

The probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

Step-by-step explanation:

Let <em>X</em> = number of individuals in the United States who held multiple jobs.

The probability that an individual holds multiple jobs is, <em>p</em> = 0.13.

The sample of employed individuals selected is of size, <em>n</em> = 225.

An individual holding multiple jobs is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 225 and <em>p</em> = 0.13.

But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion <em>p</em>.

Conditions of Normal approximation to Binomial are:

  • np ≥ 10
  • n (1 - p) ≥ 10

Check the conditions as follows:

np=225\times 0.13=29.25>10\\n(1-p)=225\times (1-0.13)=195.75>10

The distribution of the proportion of individuals who hold multiple jobs is,

p\sim N(p, \frac{p(1-p)}{n})

Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:

P(p

*Use a <em>z</em>-table.

Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

7 0
3 years ago
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