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) =
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
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