Suppose X is a normal random variable with mean 𝜇 = 70 and standard deviation 𝜎 = 5. Find a such that P(X ≥ a) = 0.01.(Round your answer to one decimal place.)

Question

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Suppose X is a normal random variable with mean 𝜇 = 70 and standard deviation 𝜎 = 5. Find a such that P(X ≥ a) = 0.01. (Round your answer to one decimal place.)

a =

Answer 1

P(x>85) = .01/2 =.005, if mean = 70 and standard deviation= 5

the empirical rule states that 99% of the data fall within 3 standard deviations from the mean (85-70)/5 = 3 = z score. Still, .005 is not .001 there's a more precise empirical rule that says 99.7% fall within 3 standard deviations from the mean. That means P(x>85) = (1-0.997)/2 = 0.003/2 = 0.0015, which is very close to 0.001. But it suggests a slightly larger value of X than 85 is needed.

use z tables and the z score is either 3.1 or 3.09 where probability = 0.001

3.1 standard deviations = 3.1 x 5 =15.5. Add that to the mean to get 85.5. or use 3.09 x 5 = 15.45. Add that to 70

P(x>85.5) = 0.001 or

P(x>85.45) = 0.001. If you want to round up to one decimal, then they're the same. 85.45 rounded to one decimal is conventionally 85.5