Question

The body temperatures of adults are normally distributed with a mean of 98.6° f and a standard deviation of 0.60° f. if 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° f. 0.0228 0.8188 0.9360 0.9772

Answer 1

To find the probability that the mean body temperature of 36 adults is greater than 98.4°F, we can use the central limit theorem and calculate the z-score. The corresponding probability is approximately 0.0228.

Step-by-step explanation:

In order to find the probability that the mean body temperature of 36 adults is greater than 98.4°F, we can use the central limit theorem. The central limit theorem states that the sampling distribution of the sample mean approaches a normal distribution as the sample size increases. Given that the body temperatures of adults are normally distributed with a mean of 98.6°F and a standard deviation of 0.60°F, we can calculate the z-score for a mean body temperature of 98.4°F and then use the standard normal distribution table to find the corresponding probability.

First, calculate the standard error of the mean using the formula:

Standard error = standard deviation / square root of sample size

Standard error = 0.60°F / √36 = 0.10°F

The z-score is calculated using the formula:

z = (x - μ) / standard error

z = (98.4 - 98.6) / 0.10 = -0.2 / 0.10 = -2

Using the standard normal distribution table, we can find the probability associated with a z-score of -2. The corresponding probability is approximately 0.0228. Therefore, the probability that the mean body temperature of 36 adults is greater than 98.4°F is approximately 0.0228.

Answer 2

Here the population standard deviation is 0.60 degree F. If a sample of 36 adults is randomly selected, that results in a sample standard deviation of 0.60 degree F divided by the square root of 36: 0.10 degree F.

The probability in question is the area under the standard normal probability distribution between 98.4 degree F and infinity, and intuitively you can detect that this will be more than 0.5 (corresponding to 50%).

Convert 98.4 degrees F to a z-score, using the sample standard deviation (0.10 degree F). That z score is
98.4-98.6
z = -------------- = -0.20/0.10 = -2
0.10

We need to determine the area under the standard normal curve to the right of z=-2. Use a table of z-scores to do this, or use your calculator's built-in probability functions. My result is 98.21% (corresponding to an area of 0.9821).

With my calculator I can find this probability using the following command:

normalcdf(-2,100000,0.10).