Question

2.Assume that z-scores are normally distributed with a mean of 0 and a standard deviation of 1.

If P(z>d)=0.8689P(z>d)=0.8689, find d.

3.A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 247.5-cm and a standard deviation of 1.9-cm. For shipment, 22 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is between 247.2-cm and 248.6-cm.
P(247.2-cm < M < 248.6-cm) =

Answer 1

2)

Using standard normal table,

P(z > -1.12) = 0.8689

d = -1.12

3)

Solution :

= / n = 1.9 / 22

= P[(247.2 - 247.5) / 1.9 / 22< ( - ) / < (248.6 - 247.5) / 1.9 / 22)]

= P(-0.74 < Z < 2.72)

= P(Z < 2.72) - P(Z < -0.74)

= 0.7671