739 rounded to the nearest hundred is
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a) The estimate of the proportion of defectives when the process is in control is 0.054
b) The standard error of the proportion if the sample size is 100 is 0.0226.
c) The upper control limit is 0.1218 and the lower control limit is 0 (since LCL < 0 and p > 0, we can write LCL = 0).
<h3>What are the formulas for finding the estimate of the proportion, standard variation, and control limits?</h3>1) The estimate of the proportion of success is
p = (number of success)/(total number of samples)
I.e., p = x/N
2) The standard deviation of the proportion of success is
3) The upper and lower control limits for a control chart are:
L.C.L = p - 3
and U.C.L = p + 3
It is given that, there are 25 samples of 100 items each.
So, the total number of items i.e., the total sample size,
N = 25 × 100 = 2500
In 25 samples, a total of 135 items were found to be defective.
So, the number of defectives x = 135
a) The estimate of the proportion of defectives is p = x/N
On substituting, we get
p = 135/2500 = 0.054
b) The standard error of the proportion if the sample of size 100 is calculated by
On substituting p = 0.054 and n = 100, we get
= 0.0226
c) The control limits for the control chart are:
Upper control limit = p + 3
⇒ U.C.L = 0.054 + 3(0.0226) = 0.054 + 0.0678 = 0.1218
Lower control limit = p - 3
⇒ L.C.L = 0.054 - 3(0.0226) = 0.054 - 0.0678 = - 0.0138 ≈ 0
(Since we know that the lower control limit should not be a negative value, it is made equal to 0).
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Answer:
20. AB = 42
21. BC = 28
22. AC = 70
23. BC = 20.4
24. FH = 48
25. DE = 10, EF = 10, DF = 20
Step-by-step explanation:
✍️Given:
AB = 2x + 7
BC = 28
AC = 4x,
20. Assuming B is between A and C, thus:
AB + BC = AC (Segment Addition Postulate)
2x + 7 + 28 = 4x (substitution)
Collect like terms
2x + 35 = 4x
35 = 4x - 2x
35 = 2x
Divide both side by 2
17.5 = x
AB = 2x + 7
Plug in the value of x
AB = 2(17.5) + 7 = 42
21. BC = 28 (given)
22. AC = 4x
Plug in the value of x
AC = 4(17.5) = 70
✍️Given:
AC = 35 and AB = 14.6.
Assuming B is between A and C, thus:
23. AB + BC = AC (Segment Addition Postulate)
14.6 + BC = 35 (Substitution)
Subtract 14.6 from each side
BC = 35 - 14.6
BC = 20.4
24. FH = 7x + 6
FG = 4x
GH = 24
FG + GH = FH (Segment Addition Postulate)
(substitution)
Collect like terms
Divide both sides by -3
FH = 7x + 6
Plug in the value of x
FH = 7(6) + 6 = 48
25. DE = 5x, EF = 3x + 4
Given that E bisects DF, therefore,
DE = EF
5x = 3x + 4 (substitution)
Subtract 3x from each side
5x - 3x = 4
2x = 4
Divide both sides by 2
x = 2
DE = 5x
Plug in the value of x
DE = 5(2) = 10
EF = 3x + 4
Plug in the value of x
EF = 3(2) + 4 = 10
DF = DE + EF
DE = 10 + 10 (substitution)
DE = 20