The body temperatures of adults are normally distributed with a mean of 98.6 degree F and a standard deviation of 0.60 degree F. If 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4 degree F.
-2 z=0.0228 1-0.0228 answer 0.9772
Use z-scores:
z = (98.4 - 98.6)/(0.60/√36) = ?
Once you calculate the z-score, check a z-table for your probability. (Remember that the problem is asking to find the probability "greater than" 98.4)
Well, it's time for a little statistical clowning around! To solve this, we can use the Central Limit Theorem to approximate the distribution of the sample means. We know that the mean of the sampling distribution of the sample means will be the same as the population mean, which is 98.6°F.
Now, let's calculate the standard deviation of the sampling distribution of the sample means, also known as the standard error. The formula for the standard error is the population standard deviation divided by the square root of the sample size. In this case, the standard deviation is 0.60°F, and the sample size is 36.
So, the standard error is 0.60°F divided by the square root of 36, which is 0.10°F.
Now, we can calculate the z-score for the sample mean. The formula for the z-score is (sample mean - population mean) divided by the standard error. In this case, the sample mean we're interested in is 98.4°F, and the population mean is 98.6°F. The standard error we just calculated is 0.10°F.
So, the z-score is (98.4 - 98.6) / 0.10, which is -2.
To find the probability that the sample mean is greater than 98.4°F, we need to find the area under the standard normal curve to the right of the z-score of -2. Fortunately, I have the perfect party trick to solve this!
Using a statistical table or calculator, we find that the area to the right of a z-score of -2 is approximately 0.9772.
So, the probability that the mean body temperature of the 36 adults selected is greater than 98.4°F is approximately 0.9772.
Remember, this is just an estimation based on the Central Limit Theorem. Now, "temper" your expectations and let the statisticians continue with their analytical seriousness!
To find the probability that the mean body temperature of 36 adults is greater than 98.4 degree F, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means approaches a normal distribution with a mean equal to the population mean (μ) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (σ/√n).
In this case, the mean body temperature of adults is normally distributed with a mean (μ) of 98.6 degree F and a standard deviation (σ) of 0.60 degree F. The sample size (n) is 36.
To calculate the probability, we need to standardize the value of 98.4 degree F using z-score. The z-score formula is given as (x - μ) / (σ/√n), where x is the value we want to standardize.
Plugging in the values, we get:
z = (98.4 - 98.6) / (0.60 / √36)
= (-0.2) / (0.10)
= -2
Next, we need to find the probability of a z-score being greater than -2. We can use a z-table or calculator to find this probability. Looking up the z-score of -2 in the z-table, we find that the probability is approximately 0.9772.
Therefore, the probability that the mean body temperature of 36 adults is greater than 98.4 degree F is approximately 0.9772, or 97.72%.