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Answer:
a.P<x<9.85 orx>10.15)=0.3174, Total defects=317.4
b.p=0.0026,total defects=2.6
c.Less of the items produced will be classified as defects.
Step-by-step explanation:
a.The standard score,z, is esentially x reduced by process mean then divided my process standard deviation.

Total defects=Production*Probability
=0.3174*1000

b. 
therefore:-

Defects=Probability*Production
=0.0026*1000
=2.6
c.Reducing process variation results in a significant reduction in the number of unit defects.
The completed table will have values filled into it accordingly from left to right and the downwards.
<h3>What values are required to complete the table?</h3>
The concept of relative frequencies involves expressing the frequency of occurrence of events as a fraction of a whole frequency.
Hence, it follows that the blank space on the top-leftmost corner of the table can be filled with the value; 4/5 = 0.8.
For the top-rightmost blank space in the table, the required value is; 2/754 = 0.0027.
For the bottom-leftmost blank space, the required value to fill the table is; 1/5 = 0.2.
For the bottom-rightmost blank space, the required value is; 752/754 = 0.9973.
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