In a study of 205 adults, the average heart rate was 75 beats per minute. Assume the population of heart rates is known to be approximately normal, with a standard deviation of 8 beats per minute. What does a margin of error of 1.1 for the 95% confidence interval of the average beats per minute mean? There is a 95% chance that the population mean is between 67 and 83 beats per minute. There is a 95% chance that the population mean is between 73.9 and 76.1 beats per minute. There is a 5% chance that the population mean is less than 75 beats per minute. There is a 5% chance that the population mean is more than 75 beats per minute.

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LegobuilderC LegobuilderC · Answer 1 · 2022-04-08T16:50:51+02:00

Answer:

There is a 95% chance that the population mean is between 73.9 and 76.1 beats per minute.

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andromache andromache · Answer 2 · 2022-04-13T07:00:11+02:00

There is a 95% chance that the population mean is between 73.9 and 76.1 beats per minute.

Since

The average heart rate was 75 beats per minute.

The standard deviation is 8 beats per minute

And, there is the study of 205 adults

Now the following formula is to be used

Since

[tex]x \pm z \frac{\sigma}{\sqrt{n} }[/tex]

Here

z = 1.96 at 95% confidence interval

So,

[tex]= 75 \pm 1.96 \frac{8}{\sqrt{205} } \\\\= 75 - 1.96 \frac{8}{\sqrt{205} } , 75 + 1.96 \frac{8}{\sqrt{205} }[/tex]

= 73.9 ,76.1

Hence, the above statement should be true.

Learn more about standard deviation here: https://brainly.com/question/20529928