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
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).
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