Question

If we know that the core temperatures of baboon fall roughly in a normal distribution with a mean of 100.6°F and a standard deviation of 0.86°F, answer the following.

Step 1 of 4:

What temperature would put a baboon in the 76th percentile? Include appropriate unit and round to 2 decimals.

Step 2 of 4:

What temperature would put a baboon in the bottom 20% of temperatures? Include appropriate unit and round to 2 decimals.

Step 3 of 4:

What is the probability that a baboon has a body temperature of 100°F or more?

Step 4 of 4:

What is the probability that ababoon has a body temperature less than 99°F?

Answer 1

We have given that,

Temperature of baboon ~ N(100.6,0.86)

Mean = 100.6

Standard deviation = 0.86

Step 1 :

We have given 76th percentile that means 76% data lies below 76th percentile and 24% data lies above it.

So below probability = 76% = 0.76

Excel function to find the temperature would put a baboon in the 76th percentile.

=NORM.INV(probability,mean,standard deviation)

=NORM.INV(0.76,100.6,0.86)

So the 101.2074= ( round to 2 decimal) temperature would put a baboon in the 76th percentile.

Step 2:

Excel function to find temperature would put a baboon in the bottom 20% of temperatures

=NORM.INV(probability, mean, standard deviation)

= NORM.INV(0.20,100.6,0.86)

So the 99.87621 = temperature would put a baboon in the bottom 20% of temperature.

Step 3 :

To find Probability that baboon has a body temperature of or more.

Excel function to find    is

=NORM.DIST(x,mean, standard deviation,TRUE)

=NORM.DIST(100,100.6,0.86,TRUE)

So

Then P(X>100) = 1- 0.24269 = 0.75731 = 0.76

So the probability that baboon has a body temperature of or more is 0.76.

Step 4 :

Excel function to find the probability that baboon has a body temperature less than .

=NORM.DIST(x,mean, standard deviation,TRUE)

=NORM.DIST( 99,100.6,0.86,TRUE)

So the probability that baboon has a body temperature less than is 0.03141 = 0.03