Question

The body temperatures of a group of healthy adults have a​ bell-shaped distribution with a mean of 98.38°F and a standard deviation of 0.48°F. Using the empirical​ rule, find each approximate percentage below.

a. What is the approximate percentage of healthy adults with body temperatures within 3 standard deviations of the​ mean, or between 97.05°F and 99.57°​F?
b. What is the approximate percentage of healthy adults with body temperatures between 97.89°F and 98.73°​F?

Answer 1

a) The problem says that this represent the values within 3 deviations from the mean and using the empirical rule we know that on this case we have 68% of the data on this interval.

b) For this case we can use the z score formula again:

`z_1= (98.73-98.38)/(0.48)=0.729`

`z_2= (97.89-98.38)/(0.48)=-1.020`

For this case we want this probability:

`P(97.89<X<98.73) =P(-1.02<Z<0.729)= P(Z<0.729)-P(Z<-1.02)=0.767-0.154= 0.613`

So the approximate percentage of temperatures between 97.89F and 98.73F is 61.3%

Explanation:

The empirical rule, also referred to as "the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)". The empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

For this case we know that the body temperatures for a group of heatlhy adults represented with the random variable X follows this distribution:

`X \sim N(\mu =98.38F, \sigma=0.48 F)`

Part a

For this case we can use the z score formula to measure how many deviations we are within the mean, given by:

`z=(x-\mu)/(\sigma)`

If we find the z score for the values given we got:

`z_1= (99.57-98.38)/(0.48)=2.479`

`z_2= (97.05-98.38)/(0.48)=-2.771`

The problem says that this represent the values within 3 deviations from the mean and using the empirical rule we know that on this case we have 68% of the data on this interval.

Part b

For this case we can use the z score formula again:

`z_1= (98.73-98.38)/(0.48)=0.729`

`z_2= (97.89-98.38)/(0.48)=-1.020`

For this case we want this probability:

`P(97.89<X<98.73) =P(-1.02<Z<0.729)= P(Z<0.729)-P(Z<-1.02)=0.767-0.154= 0.613`

So the approximate percentage of temperatures between 97.89F and 98.73F is 61.3%