Probability question

Question

Assume the population of human body temperatures has a mean of 98.6 F and a standard deviation of 0.62 F. Suppose we randomly select a sample size of n= 106. What is the probability the mean body temperature is no more than 98.2 F?

Answer 1

To compute this probability we are going to assume a normal distribution and utilize the standard normal probability tables.

The z-statistic for the sample mean = (sample mean - population mean)/standard error of the sample mean.

The standard error of the sample mean = population standard deviation/sqrt(sample size) = 0.62/sqrt(106) = 0.0602.

We are trying to find the probability the mean body temperature is no more than 98.2 which is the same thing as saying the probability the mean body temperature is less than 98.2.

We want to compute P(sample mean < 98.2). To use normal probability tables we need to standardize the sample mean by offsetting it by the population mean and dividing by the standard error of the sample mean:

P(sample mean - population mean < 98.2 - 98.6)

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standard error of sample mean 0.0602

The left side of the inequality is our z-statistic. The right side simplifies to -6.645 so that we want to compute:

P(Z < -6.645).

Using the standardized normal table or ti84 calculator (normcdf(-1000,-6.645,0,1) we determine the probability is approximately zero.

We could utilize the ti84 calculator to compute the probability without the intermediate z-statistic calculation:

normcdf(-1000,98.2,98.6,0.0602)