Answer :
1. The physicians should use temperature of 99.2 degrees. 2. F is approximately 0.002 or 0.2%. 3. The probability that sample mean of 4 individuals will exceed 98.7 degrees F is approximately 0.053 or 5.3%.
Using a standard normal distribution table or a calculator with normal distribution function. Then, use formula:
[tex]z = (x - \mu) / \sigma[/tex]
Solving for x, we get:
[tex]x = z * \sigma + \mu[/tex]
[tex]x = 1.645 * 0.62 + 98.2[/tex]
x ≈ 99.2
2. We can use same formula as in 1, but this time we want to find probability.
[tex]z = (x - \mu) / \sigma[/tex]
Solving for z:
[tex]z = (100 - 98.2) / 0.62[/tex]
z ≈ 2.90
We can find probability that z-score is greater than 2.90, which is approximately 0.002.
3. Mean of sample means = μ = 98.2
standard deviation of sample means = [tex]\sigma / sqrt(n) = 0.62 / sqrt(4) = 0.31[/tex]
Using the standard normal distribution table:
[tex]z = (98.7 - 98.2) / 0.31[/tex]
z ≈ 1.61
To know more about standard normal distribution, here
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