To find the probability of getting a sample mean temperature of 98.2°F or lower, given that the population mean (μ) is 98.6°F and the population standard deviation (σ) is 0.62°F, we can use the Central Limit Theorem.
The Central Limit Theorem states that for a sufficiently large sample size, the sampling distribution of the sample mean will be normally distributed with mean \( \mu \) and standard deviation \( \frac{\sigma}{\sqrt{n}} \).
- Calculate the standard error (SE) of the mean:
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.62}{\sqrt{106}} \approx \frac{0.62}{10.2956} \approx 0.0602 \]
- Calculate the z-score for a sample mean of 98.2°F:
\[ z = \frac{\bar{x} - \mu}{SE} = \frac{98.2 - 98.6}{0.0602} = \frac{-0.4}{0.0602} \approx -6.64 \]
- Using the z-score, find the probability corresponding to \( z = -6.64 \).
A z-score of -6.64 is very extreme and falls far into the left tail of the normal distribution. The probability of obtaining a z-score of -6.64 or lower is virtually zero.
Using standard normal distribution tables or calculators, we find that:
\[ P(Z < -6.64) \approx 0.0000 \]
Thus, the answer is approximately:
0.0001 (since this is the closest option provided).