Question

Assigned Media Question Help According to a random sample taken at 12​ A.M., body temperatures of healthy adults have a​ bell-shaped distribution with a mean of 98.3298.32degrees°F and a standard deviation of 0.590.59degrees°F. Using​ Chebyshev's theorem, what do we know about the percentage of healthy adults with body temperatures that are within 22 standard deviations of the​ mean? What are the minimum and maximum possible body temperatures that are within 22 standard deviations of the​ mean?

Answer 1

Answer:

$1- \frac{1}{2^2} = 1-\frac{1}{4} = \frac{3}{4}$

So then we will have at least 75% of the data within two deviations from the mean $\bar x \pm 2s$.

$\bar x -2s = 98.32 -2*0.59 = 97.14$

$\bar x +2s = 98.32 +2*0.59 = 99.5$

So then the interval would be (97.14, 99.5)

Explanation:

Previous concepts

Chebyshev’s rule is appropriate for any distribution. "Chebyshev’s inequality applies to all distributions, regardless of shape". And is useful since provides a "minimum percentage of the observations that lies within k standard deviations of the mean.  "

If k = 2, at least 3/4 of the measurements lie within 2 standard deviations to within the mean.

And the general formula is (1-1/k^2) represent the fraction of the data within the mean .

Solution to the problem

For this case we want to find the percentage of data that would be at least within two deviations from the mean so for this case the value of k =2 and if we replace we got:

$1- \frac{1}{2^2} = 1-\frac{1}{4} = \frac{3}{4}$

So then we will have at least 75% of the data within two deviations from the mean $\bar x \pm 2s$.

For the other part we have the mean and deviation provided $\bar x = 98.32$ $s = 0.59$ the interval would be:

$\bar x -2s = 98.32 -2*0.59 = 97.14$

$\bar x +2s = 98.32 +2*0.59 = 99.5$

So then the interval would be (97.14, 99.5)