Suppose that engineering specifications on the shelf depth of a certain slug to be turned on a cnc lathe are from .0275 in. to .0278 in. and that values of this dimension produced on the lathe can be described using a normal distribution with mean μ and standard deviation σ . (a) if μ = .0276 and σ = .0001, about what frac- tion of shelf depths are in specifications? (b) what machine precision (as measured by σ ) would be required in order to produce about 98% of shelf depths within engineering spec- ifications (assuming that μ is at the midpoint of the specifications)?

0.8188 or 82%

0.0000643

Step-by-step explanation:

Given that engineering specifications on the shelf depth of a certain slug to be turned on a CNC lathe are from .0275 in. to .0278 in.

If x is the dimension then X is N(0.0276, 0.0001)

Or is Normal (0,1)

a) Fraction  of shelf depths are in specifications

=

b) For 98% to be within the specifications,

Margin of error =

For 98% critical value Z two tailed is 2.33

Hence std error =

So sigma should be = 0.0000643

$27.5

step-by-step explanation:

$110.00 divided by 4

14 ft 7 inches

step-by-step explanation:

since each model inch corresponds to 25 real inches, 7 model inches will correspond to

7×25" = 175"

at 12" per foot, that is 14 feet, 7 inches.