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

10

step-by-step explanation:

the homeowners insurance cost for a 115,000 wood frame house is 80,500.

step-by-step explanation:

it is given that the coverage rate for wood frame house is 0.7.

the value of wood frame house is 115,000.

.

homeowners insurance cost for a 115,000 wood frame house is

therefore the homeowners insurance cost for a 115,000 wood frame house is 80,500.