6.49 Normal distribution problem.
(a) The probability of the weight being less than 7.9 is the area under the curve to the left of x = 7.9. For this value of x, . From the nomral distribution table, for z = 3.0, the area under the curve (from the mean to the value of interest) is 0.4987. The area to the left of x = 7.9 is thus 0.5 – 0.4987 = 0.0013. So 0.13% of the cans will be rejected.
(b) For this case, . For this value of z, the area under the normal distribution curve from the mean to x=7.9 is 0.4772. So the area to the left of 7.9 is 0.5 – 0.4772 = 0.0228. 2.28% of the cans will be rejected.
(c) for this case, The table does not give a value for z this high but for 4.9, it is given as 0.5. So the area to the left of 7.9 is 0.5 – 0.5 = 0. For this z value the number of rejected cans is negligible. (actually 3x10-5%.