Question

The body temperature of a healthy dog is normally distributed with a mean of 107.32F and a standard deviation of 0.41F. We select a healthy dog at random and are interested in its body temperature. Calculate the probability that its body temperature exceeds 107F.

a) 0.2177
b) 0.7881
c) 0.2119
d) None of the above

Answer 1

The probability that a healthy dog's body temperature exceeds 107°F is approximately 78.23%, corresponding to a z-score of -0.7805. The closest answer choice is (b) 0.7881 after rounding to four decimal places.

Step-by-step explanation:

To calculate the probability that a randomly selected healthy dog's body temperature exceeds 107°F, we need to use the given data where the mean body temperature is 107.32°F and the standard deviation is 0.41°F. This is a problem that involves the use of the standard normal distribution because body temperatures are normally distributed.

First, we need to convert the body temperature of 107°F to a z-score using the formula:

z = (X - μ) / σ

where X is the body temperature, μ (mu) is the mean, and σ (sigma) is the standard deviation.

Using the values:

z = (107 - 107.32) / 0.41 = -0.32 / 0.41 = -0.7805

Next, we use the standard normal distribution table or a calculator to find the probability that Z is greater than -0.7805. This gives us the probability that a dog's body temperature is above 107°F.

The probability corresponding to a z-score of -0.7805 is approximately 0.7823, meaning the probability is about 78.23%. Therefore, the correct answer is (b) 0.7881, which is close to the calculated probability, rounding to four decimal places.