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The temperature of a person has a normal distribution. What is the probability that the temperature of a randomly selected person will be within 2.42 standard deviations of its mean? Provide answer with 4 or more decimal places.

User Sam Makin
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Answer:

Explanation:

If the temperature of a person follows a normal distribution, we know that approximately 95% of the observations fall within 2 standard deviations of the mean. Since we are given that we want to find the probability of the temperature being within 2.42 standard deviations of its mean, we can use the standard normal distribution and the z-score formula.

The z-score formula is given by:

z = (x - μ) / σ

where x is the observed value, μ is the mean, and σ is the standard deviation. In this case, we want to find the probability that the temperature is within 2.42 standard deviations of the mean, so we can set:

z = 2.42

Since the normal distribution is symmetric, we can find the area to the right of the mean (z = 0) and double it to get the total probability. Using a standard normal distribution table or calculator, we find that the area to the right of z = 2.42 is approximately 0.0074. So the area to the left of z = 2.42 is approximately 0.9926.

Doubling this area gives us the total probability:

P(z < 2.42 or z > -2.42) = 2 * P(z < 2.42) = 2 * 0.9926 = 0.9852

Therefore, the probability that the temperature of a randomly selected person will be within 2.42 standard deviations of its mean is 0.9852, or approximately 0.9852 with four decimal places.

User Elephantik
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