In: Statistics and Probability
Let X ∼ Normal(µ, 1).
(a) Give an interval (L, U), where U and L are based on X, such that P(L < µ < U) = 0.99.
(b) Give an upper bound U based on X such that P(µ < U) = 0.99.
(c) Give a lower bound L based on X such that P(L < µ) = 0.99.
We are given that:
X ∼ N(µ, 1) => Z = X - µ ~ N(0,1), the standard normal distribution.
(a)
From the table of standard normal distribution, we get:
P(-2.575829 < Z < 2.575829) = 0.99
=> P(-2.575829 < X - µ < 2.575829) = 0.99
=> P(-2.575829 < µ - X < 2.575829) = 0.99
=> P(-2.575829 + X < µ < 2.575829 + X) = 0.99
=> (L = -2.575829 + X, U = 2.575829 + X) is the desired interval such that P(L < µ < U) = 0.99. [ANSWER]
(b)
From the table of standard normal distribution, we get:
P(Z > -2.326348) = 0.99
=> P(X - µ > -2.326348) = 0.99
=> P(µ - X < 2.326348) = 0.99
=> P(µ < 2.326348 + X) = 0.99
=> U = 2.326348 + X is an upper bound based on X such that P(µ < U) = 0.99. [ANSWER]
(c)
From the table of standard normal distribution, we get:
P(Z < 2.326348) = 0.99
=> P(X - µ < 2.326348) = 0.99
=> P(µ - X > -2.326348) = 0.99
=> P(µ > -2.326348 + X) = 0.99
=> L = -2.326348 + X is a lower bound based on X such that P(L < µ) = 0.99. [ANSWER]
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