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If you select a person at random from a population with body temperatures that are N(98.2, 0.62)...

a. 0.2119
b. 0.7881
c. 0.5000
d. 0.2898

User Ncatnow
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1 Answer

3 votes

Final answer:

The probability of selecting a person with a temperature greater than 98.6 degrees Fahrenheit from a population with normally distributed body temperatures is 0.5000.

Step-by-step explanation:

The question asks for the probability of selecting a person at random from a population with body temperatures that are normally distributed with a mean of 98.2 and a standard deviation of 0.62. To find this probability, we need to calculate the z-score and look it up in the standard normal distribution table.

First, we calculate the z-score: z = (x - mu) / sigma, where x is the body temperature, mu is the mean, and sigma is the standard deviation. In this case, x is 98.6, mu is 98.2, and sigma is 0.62. Plugging in the values, we get z = (98.6 - 98.2) / 0.62 = 0.64.

Next, we look up the z-score in the standard normal distribution table. The closest value to 0.64 is 0.7389. Since we're looking for the probability of selecting a person with a temperature greater than 98.6, we subtract the probability of selecting a person with a temperature less than or equal to 98.6 from 0.5 (since the distribution is symmetric).

Therefore, the answer is (c) 0.5000.

User Raja Hammad Farooq
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