Question

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.

Answer 1

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.

Answer 2

You can use the Chebyshev's inequality/Theorem to find the needed probability and percentage.

The answers are

The minimum and maximum possible body temperatures that are within 2 standard deviations of the mean is given as

Minimum temperature = 97.04˚F

Maximum temperature = 99.28˚F

At least ___75____% of healthy adults have temperatures within 2 standard deviations of 98.16˚F(its mean).

What is Chebyshev's Inequality/Theorem?

If a random variable X is integrable and has mean(or say expectation) $\mu$ (finite) and standard deviation $\sigma$, then for any real number k >0, we have

$P(|X - \mu} | \geq k\sigma) = \dfrac{1}{k^2}\\\\or\\\\P(|X - \mu| \leq k\sigma) = 1 - \dfrac{1}{k^2}$

(take notice of the inequality sign)

Using the above theorem for given condition to find the needed information

Let the random variable be X which tracks the body temperature.

Then we have:

E(X) = expected value of X = $\mu$ = 98.16˚F

Standard deviation of X  = $\sigma$ = 0.56˚F

The minimum temperature and maximum temperature within  2 standard deviations of the mean is

$|X - \mu| \leq 2 \sigma\\-2\sigma \leq X - \mu \leq 2 \sigma\\ \mu-2\sigma \leq X \leq \mu + 2 \sigma\\\\X = 98.16 + 2 \times 0.56 = 99.28^\circ F\\X = 98.16 -2 \times 0.56 = 97.04^\circ F$

Thus, minimum temperature = 97.04˚F

maximum temperature = 99.28˚F

Using Chebyshev's inequality, we get:

$P(|X - \mu| \leq k\sigma) = 1 - \dfrac{1}{k^2}\\\\P(|X - 98.16| \leq 2 \times 0.56) = 1 - \dfrac{1}{2^2} = \dfrac{3}{4} = 0.75$    = 75% temperature will lie within 2 standard deviations of the mean. (probability is count per one, and percent is count per 100, thus, multiplying 100 to probability gives percent)

Thus,

The minimum and maximum possible body temperatures that are within 2 standard deviations of the mean is given as

Minimum temperature = 97.04˚F

Maximum temperature = 99.28˚F

At least ___75____% of healthy adults have temperatures within 2 standard deviations of 98.16˚F(its mean).

Learn more about Chebyshev's inequality/theorem here:

https://brainly.com/question/14886758