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Alinara [238K]
2 years ago
15

An insurance company reported that, on average, claims for a certain medical procedure are $942. An independent organization con

structed a 95% confidence interval of ($930 , $950) for the average amount claimed for the particular medical procedure. What conclusion best evaluates the truthfulness of the number reported by the insurance company?
Mathematics
1 answer:
kirza4 [7]2 years ago
7 0

Answer:

Step-by-step explanation:

The question is incomplete. The complete question is:

An insurance company reported that, on average claims for a certain medical procedure are $942. an independent organization constructed a 95% confidence interval of ($930, $950) for the average amount claimed for the particular medical procedure. what conclusion best evaluates the truthfulness of the number reported by the insurance company?

a) with 95% certainty, the average claim for this medical procedure is $942.

b) with 95% certainty, the average claim for this medical procedure is not $942.

c) the confidence interval is consistent with an average claim of $942 for this medical procedure

Solution:

Confidence interval is used to express how confident we are that the population parameter that we are looking for is contained in a range of given values. Looking at the given confident interval, the lower limit is $930 and the upper limit is $950. We can see that the population mean, $942 lies within these values. The correct option would be

c) the confidence interval is consistent with an average claim of $942 for this medical procedure

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Aleks04 [339]

The markup percentage is 45.14%

Step-by-step explanation:

The given is:

  • The selling price of a box of crackers is $1.75
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We need to find the markup percentage

The markup percentage = \frac{New-Old}{old} × 100%

∵ The selling price of a box of crackers is $1.75

∴ Old = 1.75

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∴ New = 2.54

- Substitute these values in the rule above

∵ The markup percentage = \frac{2.54-1.75}{1.75} × 100%

∴ The markup percentage = \frac{0.79}{1.75} × 100%

∴ The markup percentage = 0.4514 × 100%

∴ The markup percentage = 45.14%

The markup percentage is 45.14%

Learn more:

You can learn more about percentage in brainly.com/question/1834017

#LearnwithBrainly

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alukav5142 [94]

Answer:

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Step-by-step explanation:

We are given the following information in the question:

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We are given that the distribution of tutoring prices is a bell shaped distribution that is a normal distribution.

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z_{score} = \displaystyle\frac{x-\mu}{\sigma}

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