Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 195.3 cm and a standard deviation of 1 cm. For shipment, 24 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 between 194.8 cm and 194.9 cm.


Enter your answer as a number accurate to 4 decimal places.

Answer 1

Solution :

Given that ,

mean = = 195.3

standard deviation = = 1

n = 24

= 195.3

=  / n= 1/ 24=0.20

P(194.8<    <194.9 ) = P[(194.8-195.3) / 0.20< ( - ) / < (194.9-195.3) / 0.20)]

= P( -2.5< Z <-2 )

= P(Z <-2 ) - P(Z <-2.5 )

Using z table

=0.0228-0.0062

=0.0166

probability= 0.0166