Suppose the average body temp is 98.6 assume the variable is normally distributed with a standard deviation of .74 degrees. suppose a sample of 100 is taken. what is the probability that the mean is below 98.4??? Between 98.5 AND 98.7??

z = (98.5-98.6)/.74/sqr(100)

z = (98.7- 98.6)/.74/sqr(100)

To find the probability of certain sample means, we can use the concept of standard deviation and the assumption that the variable is normally distributed.

Step 1: Calculate the standard error of the mean (SEM).
The standard error of the mean can be calculated using the formula: SEM = standard deviation / square root of the sample size.
In this case, the standard deviation is 0.74 degrees and the sample size is 100.
SEM = 0.74 / √100 = 0.074

Step 2: Convert the given means to z-scores.
To use the z-score formula, we need to convert the given means to z-scores (standardized scores).
The formula for z-score is: z = (x - μ) / SEM, where x is the given mean and μ is the population mean.
For the first scenario (mean below 98.4):
z1 = (98.4 - 98.6) / 0.074 = -0.02 / 0.074 = -0.27

Step 3: Find the corresponding probabilities.
Using a z-table or software, look up the probabilities corresponding to the calculated z-scores.

For the first scenario (mean below 98.4):
P(mean < 98.4) = P(z < -0.27)
By referring to the z-table, we find that the probability of a z-score less than -0.27 is approximately 0.3944.

For the second scenario (mean between 98.5 and 98.7):
z2 = (98.5 - 98.6) / 0.074 = -0.1 / 0.074 = -1.35
z3 = (98.7 - 98.6) / 0.074 = 0.1 / 0.074 = 1.35
P(98.5 < mean < 98.7) = P(-1.35 < z < 1.35)
Again, by referring to the z-table, we find that the probability of z-score between -1.35 and 1.35 is approximately 0.8643.

So, the probabilities are as follows:
- P(mean < 98.4) = 0.3944 (or approximately 39.44%)
- P(98.5 < mean < 98.7) = 0.8643 (or approximately 86.43%)

Note: These calculations assume a normal distribution and may vary slightly depending on the z-table used or software for calculations.