Question

Assume a normal distribution and find the following probabilities.

(Round the values of z to 2 decimal places. Round your answers to 4 decimal places.)

(a) P(x < 17 | μ = 21 and σ = 3)

(b) P(x ≥ 72 | μ = 60 and σ = 9)

(c) P(x > 55 | μ = 60 and σ = 5)

(d) P(14 < x < 22 | μ = 19 and σ = 3)

(e) P(x ≥ 93 | μ = 80 and σ = 1.83)

Answer 1

(a) P(x < 17 | μ = 21 and σ = 3) = 0.0912
(b) P(x ≥ 72 | μ = 60 and σ = 9) = 0.164
(c) P(x > 55 | μ = 60 and σ = 5) = 0.242
(d) P(14 < x < 22 | μ = 19 and σ = 3) = 0.7935
(e) P(x ≥ 93 | μ = 80 and σ = 1.83) = 0

Solution: Using standard normal

(a) P(x < 17 | μ = 21 and σ = 3)=p(Z<(17-21)/3) = p(Z <-1.33) = 0.0912

(b) P(x ≥ 72 | μ = 60 and σ = 9) = p(Z ≥(72-60)/9) = p(Z ≥1.33) = 0.164

(c) P(x > 55 | μ = 60 and σ = 5) = p(Z>(55-60)/5) = p(Z >-1) = 0.242

(d) P(14 < x < 22 | μ = 19 and σ = 3) =P( x < 22 | μ = 19 and σ = 3) - P( x < 14 | μ = 19 and σ = 3)

P( x < 22 | μ = 19 and σ = 3) = p(Z<(14-19)/3) = p(Z <-1.67) = 0.0478
P( x < 14 | μ = 19 and σ = 3) = p(Z<(22-19)/3) = p(Z <1) = 0.8413

P( x < 22 | μ = 19 and σ = 3) - P( x < 14 | μ = 19 and σ = 3) =0.8413-0.0478 = 0.7935

(e) P(x ≥ 93 | μ = 80 and σ = 1.83) = p(Z ≥(93-80)/1.83) = p(Z ≥7.10) = 0