Human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.62°F if 19 people are randomly selected find the probability that they're mean by temperature will be less than 98.50 degrees Fahrenheit

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To find the probability that the mean body temperature of a sample of 19 people will be less than 98.50 degrees Fahrenheit, we need to use the concept of the sampling distribution of the sample mean.

1. Calculate the standard deviation of the sample mean (also known as the standard error). This is equal to the standard deviation of the population divided by the square root of the sample size.

Standard Error = standard deviation / √(sample size)
= 0.62°F / √19
≈ 0.142°F

2. Calculate the z-score, which measures how many standard errors the sample mean is away from the population mean.

Z-score = (sample mean - population mean) / standard error
= (98.50°F - 98.20°F) / 0.142°F
≈ 2.11

3. Use a z-table or a statistical software to find the probability corresponding to the calculated z-score. The probability of getting a mean body temperature less than 98.50°F can be found by looking up the z-score on the z-table.

P(Z < 2.11)

The z-table will give you the probability corresponding to the calculated z-score. For example, P(Z < 2.11) is approximately 0.982.

Therefore, the probability that the mean body temperature of the sample of 19 people will be less than 98.50 degrees Fahrenheit is approximately 0.982 or 98.2%.