Answer:
The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.
Step-by-step explanation:
We consider "lack of control by at least the third point plotted" if at least one of the three first points is over the UCL or under the LCL.
The probability of one point of being over UCL=104 is:
![z=(X-\mu)/\sigma=(104-98)/8=6/8=0.75\\\\P(X>104)=P(z>0.75)=0.227](https://tex.z-dn.net/?f=z%3D%28X-%5Cmu%29%2F%5Csigma%3D%28104-98%29%2F8%3D6%2F8%3D0.75%5C%5C%5C%5CP%28X%3E104%29%3DP%28z%3E0.75%29%3D0.227)
The probability of one point of being under LCL=96 is:
![z=(X-\mu)/\sigma=(96-98)/8=-2/8=-0.25\\\\P(X](https://tex.z-dn.net/?f=z%3D%28X-%5Cmu%29%2F%5Csigma%3D%2896-98%29%2F8%3D-2%2F8%3D-0.25%5C%5C%5C%5CP%28X%3C96%29%3DP%28z%3C-0.25%29%3D0.401)
Then, the probability of exhibit lack of control is:
![P=P(X>104)+P(X](https://tex.z-dn.net/?f=P%3DP%28X%3E104%29%2BP%28X%3C96%29%3D0.227%2B0.401%3D0.628)
The probability of having at least one point out of control in the first three points is:
![P(x\geq1)=1-P(0)\\\\P(x\geq1)=1- 0.052=0.948\\\\\\P(x=0) = \binom{11}{0} p^{0}q^{3}= 1*1*0.0515=0.0515\\\\](https://tex.z-dn.net/?f=P%28x%5Cgeq1%29%3D1-P%280%29%5C%5C%5C%5CP%28x%5Cgeq1%29%3D1-%200.052%3D0.948%5C%5C%5C%5C%5C%5CP%28x%3D0%29%20%3D%20%5Cbinom%7B11%7D%7B0%7D%20p%5E%7B0%7Dq%5E%7B3%7D%3D%201%2A1%2A0.0515%3D0.0515%5C%5C%5C%5C)
The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.
The answer is definitelly <span>C.(Area of 1) + (Area of 2) = (Area of 3)</span>
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Answer:
coordinates of the Midpoint = (7 ,4).
Step-by-step explanation:
Given : endpoints are H(10, 1) and K(4, 7).
To find :Find the coordinates of the midpoint .
Solution : We have given that endpoints are H(10, 1) and K(4, 7).
By the mid point segment formula :
.
Here,
Md point =
.
Mid point =
.
( 7 ,4)
coordinates of the Midpoint = (7 ,4).
Therefore, coordinates of the Midpoint = (7 ,4).
Solve the system of linear equations. 9x-19y=74 and 3x+3y=6<span>3x+3y=6
Solve equation #2 for x
3x = 6-3y
x = 2-y
</span>Plug x above in equation #1 to find y
<span>9x-19y=74
</span><span>9(2-y) - 19y = 74
18 - 9y -19y = 74
-28y = 74 - 18
-28y =56
y =56/(-28)
y = -2
plug y = -2 into </span>x = 2-y to look for x
<span>
x = 2 -(-2)
x = 2 +2
x = 4
OK answer x = 4 and y = -2
{x, y} = {4, -2}
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