Question
Assume that the population of human body temperatures has a mean of 98.6°F, as is commonly believed. Also assume that the population standard deviation is 0.62°F. If a sample of size n = 106 is randomly selected, find the probability of getting a mean temperature of 98.2°F or lower.
(1 point)
2
0.0001
0.9999
0.2578
0.4800
(1 point)
2
0.0001
0.9999
0.2578
0.4800
Answers
There are no human answers yet.
Answered by
GPT-4o mini
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 \( \frac{\sigma}{\sqrt{n}} \).
1. Calculate the standard error (SE) of the mean:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{0.62}{\sqrt{106}} \approx \frac{0.62}{10.2956} \approx 0.0602
\]
2. Calculate the z-score for a sample mean of 98.2°F:
\[
z = \frac{\bar{x} - \mu}{SE} = \frac{98.2 - 98.6}{0.0602} = \frac{-0.4}{0.0602} \approx -6.64
\]
3. 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).
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 \( \frac{\sigma}{\sqrt{n}} \).
1. Calculate the standard error (SE) of the mean:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{0.62}{\sqrt{106}} \approx \frac{0.62}{10.2956} \approx 0.0602
\]
2. Calculate the z-score for a sample mean of 98.2°F:
\[
z = \frac{\bar{x} - \mu}{SE} = \frac{98.2 - 98.6}{0.0602} = \frac{-0.4}{0.0602} \approx -6.64
\]
3. 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).
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.