This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: and 0.9030 cm. A sample of 10 shafts was taken from a shaftproducing machine. Their diameters turned out to be: 0.9030, 0.9032, 0.9031, 0.9039, 0.9030, 0.9038, 0.9002, 0.9020, 0.9018, 0.8996. a) Calculate an unbiased estimate of the proportion of inspec shafts from this sample. b) Calculate unbiased estimates of the mean and variance of the shaft diameter produced by the machine. c) Assuming that the distribution of shaft diameters is normal, and that it is impossible to change the variance, calculate the percentage of inspec shafts that the machine would produce if we adjusted the mean optimally. 4. Suppose a 1700 insurer wants to pin down the probability of a ship 1 being lost with a precision of 0.01 with probability of 99%. How many past cases does he need to look at in the worst case? What is the answer if he believes that the probability is no more than 0.03? 2...
View
Full Document
 Spring '08
 Perevalov
 Probability theory, Normal approximation, 99%, 0.9030 cm

Click to edit the document details