Respuesta :
Answer:
The probability that the average length of a randomly selected bundle of steel rods is between 155.6-cm and 156.2-cm
P(155.6 < x⁻ < 156.2) = 0.174 cm
Step-by-step explanation:
Given sample size 'n' = 11 steel rods
Mean of the Population = 155.1 cm
Standard deviation of the Population = 2.2 cm
Given x⁻ be the random variable of Normal distribution
Let x₁⁻ = 155.6 cm
[tex]Z_{1} = \frac{x^{-} _{1}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{155.6-155.1}{\frac{2.2}{\sqrt{11} } } = 0.7541[/tex]
Let x₂⁻ = 156.2 cm
[tex]Z_{2} = \frac{x^{-} _{2}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{156.2-155.1}{\frac{2.2}{\sqrt{11} } } = 1.659[/tex]
The probability that the average length of a randomly selected bundle of steel rods is between 155.6-cm and 156.2-cm.
P(x⁻₁ < x⁻ <x⁻₂) = P(Z₁ < Z <Z₂)
= P(Z <Z₂) - P(Z<Z₁)
= 0.5 +A(1.629) - (0.5 +A(0.7541)
= A(1.629) - A(0.7541)
= 0.4474 - 0.2734
= 0.174
Conclusion:-
The probability that the average length of a randomly selected bundle of steel rods is between 155.6-cm and 156.2-cm = 0.174
