According to a random sample taken at 12 a. m., body temperatures of healthy adults have a bell-shaped distribution with a mean of 98.16˚f and a standard deviation of 0.56˚f. using chebyshev's theorem, what do we know about the percentage of healthy adults with body temperatures that are within 2standard deviations of the mean? what are the minimum and maximum possible body temperatures that are within 2standard deviations of the mean? at least %of healthy adults have body temperatures within 2standard deviations of 98.16˚f.

Chebyshev’s Theorem establishes that at least 1 - 1/k²  of the population lie among k standard deviations from the mean.

This means that for k = 2,  1 - 1/4 = 0.75. In other words, 75% of the total population would be the percentage of healthy adults with body temperatures that are within 2standard deviations of the mean.

The maximum value of that range would be simply  μ + 2s, where μ is the mean and s the standard deviation. In the same way, the minimum value would be μ - 2s:

maximum = μ + 2s = 98.16˚F + 2*0.56˚F = 99.28˚F

minimum = μ - 2s = 98.16˚F - 2*0.56˚F  = 97.04˚F

In summary, at least 75% of the amount of healthy adults have a body temperature within 2 standard deviations of 98.16˚F, that is to say, a body temperature between 97.04˚F and 99.28˚F.