A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 255.9 cm and a standard deviation of 0.9 cm. For shipment, 23 steel rods are bundled together.
Note: Even though our sample size is less than 30, we can use
the z score because
1) The population is normally distributed and
2) We know the population standard deviation, sigma.
Find the probability that the average length of a randomly selected
bundle of steel rods is greater than 256 cm.
Enter your answer as a number accurate to 4 decimal places.
Solution :
Given that ,
mean = = 255.9
standard deviation = =0.9
n = 23
The population is normally distributed
_{} = 255.9
_{} = / n = 0.9 / 23=0.18766
P( <256 ) = P[( - _{} ) / _{} < (256 -255.9) / 0.18766]
= P(z <0.53 )
Using z table
=0.7019
probability=0.7019
Get Answers For Free
Most questions answered within 1 hours.