Question

Given that x is a normal variable with mean μ = 114 and standard deviation σ = 12, find the following probabilities. (Round your answers to four decimal places.)

(a)  P(x ≤ 120)


(b)  P(x ≥ 80)


(c)  P(108 ≤ x ≤ 117)

Answer 1

a)
Here, μ = 114, σ = 12 and x = 120. We need to compute P(X <= 120). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (120 - 114)/12 = 0.5

Therefore,
P(X <= 120) = P(z <= (120 - 114)/12)
= P(z <= 0.5)
= 0.6915

b)
z = (x - μ)/σ
z = (80 - 114)/12 = -2.83

Therefore,
P(X >= 80) = P(z <= (80 - 114)/12)
= P(z >= -2.83)
= 1 - 0.0023 = 0.9977

c)
z = (x - μ)/σ
z1 = (108 - 114)/12 = -0.5
z2 = (117 - 114)/12 = 0.25

Therefore, we get
P(108 <= X <= 117) = P((117 - 114)/12) <= z <= (117 - 114)/12)
= P(-0.5 <= z <= 0.25) = P(z <= 0.25) - P(z <= -0.5)
= 0.5987 - 0.3085
= 0.2902