The yield strength (ksi) for A36 steel is normally distributed with µ = 43 and σ = 4.5. (a) What is the 25th percentile of the distribution of A36 steel strength? (b) What strength value separates the...

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A Finding 25th percentile
Normal Distribution
µ = 43
σ = 4.5
X
µ = 43
σ = 4.50.25
P25
43
25th percentile is given by:
P25 = µ+ σ × invNorm(0.25)
µ = 43 , σ = 4.5 ,
For 25th percentile, invNorm(0.25) = −0.6745 (From z table , using interpola-
tion, 17/31 th distance between −0.67 and −0.68 )
P25 = µ+ σ × invNorm(0.25) = 43 + (4.5) ∗ (−0.6745) = 39.964750 ≈ 39.965
P25 = 39.965
Also, the exact answer using TI-83/84 or excel is
43 + invNorm(0.25)× 4.5 = 39.9647961241
X
B P (X > x0) = 0.1
Normal Distribution, µ = 43, σ = 4.5,
X
µ = 43
σ = 4.5
0.1
x0
43
P (X > x0) = 0.1 ⇒ P (X < x0) = 0.9
⇒ P
(
X − µ
σ
<
x0 − µ
σ
)
= 0.9 ⇒ P
(
Z <
x0 − µ
σ
)
=...