2018 in the soup?
Let Z ∼ N ( μ = 2 0 1 8 , σ 2 = 2 0 1 8 2 ) be a random variable with a normal distribution of mean μ = 2 0 1 8 and variance σ 2 = 2 0 1 8 2 .
What is the value of a such that the probability P ( Z ≤ 4 0 7 6 3 6 0 ) = P ( Z ≥ a ) ?
The answer is -4072324.
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2 solutions
P ( Z ≤ 4 0 7 6 3 6 0 ) = P ( ( Z − 2 0 1 8 ) ≤ ( 4 0 7 6 3 6 0 − 2 0 1 8 ) ) = P ( 2 0 1 8 Z − 2 0 1 8 ≤ 2 0 1 8 4 0 7 6 3 6 0 − 2 0 1 8 ) = P ( 2 0 1 8 Z − 2 0 1 8 ≤ 2 0 1 9 ) = (Due to 2 0 1 8 Z − 2 0 1 8 ∼ N ( 0 , 1 ) ) = P ( 2 0 1 8 Z − 2 0 1 8 ≥ − 2 0 1 9 ) = P ( Z − 2 0 1 8 ≥ − 2 0 1 9 ⋅ 2 0 1 8 ) = P ( Z ≥ 2 0 1 8 − 2 0 1 9 ⋅ 2 0 1 8 ) = P ( Z ≥ − 4 0 7 2 3 2 4 )
Normal distribution curves are symmetrical about the mean. Since the mean is 2018, 2 a + 4 0 7 6 3 6 0 = 2 0 1 8 ⟹ a = − 4 0 7 2 3 2 4