Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 179.5-cm and a standard deviation of 0.7-cm. For shipment, 20 steel rods are bundled together.

Find P₉₅, which is the average length separating the smallest 95% bundles from the largest 5% bundles.
P₉₅ = -cm

Enter your answer as a number accurate to 2 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 179.5-cm and a standard deviation of 0.7-cm. For shipment, 20 steel rods are bundled together.

Find P₉₅, which is the average length separating the smallest 95% bundles from the largest 5% bundles.
P₉₅ = -cm

Enter your answer as a number accurate to 2 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Standard error = sd/ sqrt(n) = 0.7/sqrt(20) = 0.156525 = 0.1565 ( four decimals)

Z value for 95^th percentile=1.645

The required x= mean+z*sd = 179.5+1.645*0.1565 =179.7574

=179.76 ( two decimals)