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

You may need to use the appropriate appendix table or technology to answer this question.

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

z = almost 2.33 corresponds to 0.01

5x2.33 = 16.5

16.5 + 70 = 86.5

Pr(x>86.5) = 0.01

but it might be lower, maybe 86.4 or 86.3. the tables don't give an exact z score for 0.01

using 2.32 gives 5x2.32 = 16, 70+16 = 86. z tables show a probability 2 times closer to 2.33 than 2.32, which means 2 times closer to 16.5 than 16, that's 16.333... add 70 to get 86.3 rounded off.

Pr(x>86.3) =0.01

a = 86.3

Answer 2

Formula for Z-score is:

Z=(x - 𝜇)/α 

The value of Z in the technology is:

Z=2.3263

(I'm assuming that it is one-tailed test)

Use the formula with the given

2.3263 = (x - 70)/5

11.6315= x-70

x = 81.6315

This means P(X≥ 81.6315) =0.01

Therefore: α = 81.6315