Question

A synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with mean 75.5 psi and standard deviation 3.5 psi. Let X = tensile strength of the synthetic fiber from a fiber specimen used in carpet manufacturing (in psi).   Suppose you randomly pick a sample of n = 36 fiber specimens and perform tensile testing on them. (round 5 decimal places)

a.) For n = 36 fiber specimens, what's  the probability that the average tensile strength of all the samples won't exceed 76 psi?

b.) For n = 36 fiber specimens, what's the probability that the average tensile strength of all the samples exceeds 76.75?

c.) For n = 36 fiber specimens, what's the probability that the average tensile strength of all the samples is between 74.5 and 76.5?

Answer 1

Part a)

X ~ N ( µ = 75.5 , σ = 3.5 )
P ( X > 76 ) = 1 - P ( X < 76 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 76 - 75.5 ) / ( 3.5 / √ ( 36 ) )
Z = 0.8571
P ( ( X - µ ) / ( σ / √ (n)) > ( 76 - 75.5 ) / ( 3.5 / √(36) )
P ( Z > 0.86 )
P ( X̅ > 76 ) = 1 - P ( Z < 0.86 )
P ( X̅ > 76 ) = 1 - 0.8043
P ( X̅ > 76 ) = 0.1957

Part b)

X ~ N ( µ = 75.5 , σ = 3.5 )
P ( X > 76.75 ) = 1 - P ( X < 76.75 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 76.75 - 75.5 ) / 3.5
Z = 0.3571
P ( ( X - µ ) / σ ) > ( 76.75 - 75.5 ) / 3.5 )
P ( Z > 0.3571 )
P ( X > 76.75 ) = 1 - P ( Z < 0.3571 )
P ( X > 76.75 ) = 1 - 0.6395
P ( X > 76.75 ) = 0.3605

Part c)

X ~ N ( µ = 75.5 , σ = 3.5 )
P ( 74.5 < X < 76.5 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 74.5 - 75.5 ) / ( 3.5 / √(36))
Z = -1.7143
Z = ( 76.5 - 75.5 ) / ( 3.5 / √(36))
Z = 1.7143
P ( -1.71 < Z < 1.71 )
P ( 74.5 < X̅ < 76.5 ) = P ( Z < 1.71 ) - P ( Z < -1.71 )
P ( 74.5 < X̅ < 76.5 ) = 0.9568 - 0.0432
P ( 74.5 < X̅ < 76.5 ) = 0.9135