A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 266.8-cm

Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 266.8-cm and a standard deviation of 2.2-cm.

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 266.8-cm and a standard deviation of 2.2-cm.

Find the probability that the length of a randomly selected steel rod is between 270.3-cm and 271.4-cm. 

P(270.3-cm < X < 271.4-cm) = 

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Hi Xavier,

This is another problem featuring the statistics classic equation:

z= (x-mu)/sigma

x= value you're interested in

mu= mean

sigma= standard deviation

Now, in this case, we are interested in the area between two x-values, so we'll designate the x and z variables as x₁, x₂, z₁, and z₂.

x₁ = 270.3

mu = 266.8

sigma = 2.2

z₁= (270.3 - 266.8)/ 2.2

z₁ = 1.59 Keep this value in mind.

x₂= 271.4

mu= 266.8

sigma = 2.2

z₂ = (271.4 - 266.8) / 2.2

z₂ = 2.09

Now, from z-table:

P(Z<1.59) = 0.9441

P(Z<2.09) = 0.9817

To get the area between, subtract the smaller from the larger:

P = 0.9817 - 0.9441

P = 0.0376

I hope this helps.