Question

A company produces steel rods. The lengths of all their steel rods are normally distributed with a mean of 155.1-cm and a standard deviation of 2.2-cm. For shipment, 11 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 155.6-cm and 156.2-cm.

Answer 1

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

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

`Z_(1) = (x^(-) _(1)-mean )/((S.D)/(√(n) ) ) = (155.6-155.1)/((2.2)/(√(11) ) ) = 0.7541`

Let x₂⁻ = 156.2 cm

`Z_(2) = (x^(-) _(2)-mean )/((S.D)/(√(n) ) ) = (156.2-155.1)/((2.2)/(√(11) ) ) = 1.659`

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