** 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}%.