red independent.Problem 13. Suppose a mechanical engineer designs a machine to punch holes inof sheet metal. The intended hole size is 1o0 mm, and the machine's hole punchingfollows a normal distribution with umachine to a manufacturing company whose tolerance for error is 2.5 mm (that is, if thehole isbe useable), Whar is tlie probability that his machine will punch hols in the sheet metalaccording to the manufacturer's tolerance?piece100mm and o2mm. IHe wants to sell theIS,punched esther 2.5 mm too big, or 2,5 mm too snall, then the sheet metal won't

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Asked Nov 19, 2019
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Can you do problem 13?

red independent.
Problem 13. Suppose a mechanical engineer designs a machine to punch holes in
of sheet metal. The intended hole size is 1o0 mm, and the machine's hole punching
follows a normal distribution with u
machine to a manufacturing company whose tolerance for error is 2.5 mm (that is, if the
hole is
be useable), Whar is tlie probability that his machine will punch hols in the sheet metal
according to the manufacturer's tolerance?
piece
100mm and o
2mm. IHe wants to sell the
IS,
punched esther 2.5 mm too big, or 2,5 mm too snall, then the sheet metal won't
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red independent. Problem 13. Suppose a mechanical engineer designs a machine to punch holes in of sheet metal. The intended hole size is 1o0 mm, and the machine's hole punching follows a normal distribution with u machine to a manufacturing company whose tolerance for error is 2.5 mm (that is, if the hole is be useable), Whar is tlie probability that his machine will punch hols in the sheet metal according to the manufacturer's tolerance? piece 100mm and o 2mm. IHe wants to sell the IS, punched esther 2.5 mm too big, or 2,5 mm too snall, then the sheet metal won't

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Expert Answer

Step 1

It is given that the machines hole-punching follows normal distribution with mean 100 and standard deviation 2.

That is, µ= 100, σ = 2.

Also, the intended hole-size is 100mm.

It is given that if the whole is...

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P(100 2.5< X<100 +2.5) P(97.5 < X <102.5) 97.5-100 = P 102.5100 <Z< 2 2 =P(-1.25Z<1.25) Use Standard = P(Z<1.25)-P(Z<-1.25)| Normal Table =0.89435-0.10565 =0.7887

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