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nikdorinn [45]
4 years ago
10

Someone answer these

Mathematics
1 answer:
aksik [14]4 years ago
3 0

Answer:

1. 8.5 x 10^8

2. 93/10000 - 4 x 10^12

3. 9.95 x 10^12

Step-by-step explanation:

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A data set includes 103 body temperatures of healthy adult humans having a mean of 98.5°F and a standard deviation of 0.61°F. Co
charle [14.2K]

Answer:

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is between 98.3ºF and 98.7ºF. 98.6°F is part of the confidence interval, which means that the sample suggests that this is a correct measure.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.99}{2} = 0.005

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.005 = 0.995, so Z = 2.575

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.575\frac{0.61}{\sqrt{103}} = 0.2

The lower end of the interval is the sample mean subtracted by M. So it is 98.5 - 0.2 = 98.3ºF.

The upper end of the interval is the sample mean added to M. So it is 98.5 + 0.2 = 98.7ºF.

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is between 98.3ºF and 98.7ºF. 98.6°F is part of the confidence interval, which means that the sample suggests that this is a correct measure.

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3 years ago
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A particular fruit's weights are normally distributed, with a mean of 651 grams and a standard deviation of 27 grams. if you pic
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