Statistics - Normal Distribution (Find the z value then write solutions)
Suppose the Production Operation Manager at the canning plant for evaporated milk is faced with a problem. Because of the designed machine, it generated a standard deviation of 7 ml in the filling of evaporated milk, but the average ml per fill can be adjusted. Under fear of revocation of license to operate and penalty charges by the Department of Trade and Industry (DTI), the management of the company has ordered the manager to set the filling machine so that there is a maximum 10% chance of a 370 ml can content less than 356 ml. What should the average be?
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Answer:
General term of expansion (a+b)ⁿ is
\bf \: T_{r+1} = \: \: ^nC_r \: \: \large \frak{ a ^{n−r} b ^r}T
r+1
=
n
C
r
a
n−r
b
r
For (x+2y)⁹,
Putting n =9, a=x, b=2y
\begin{gathered} \bf \: T_{r+1} = \: \: ^{9} C_r (x) ^{9−r} (2y) ^r \\ \\ \bf \: T _{r+1} = \: \: ^{9} C_r (x) ^{9−r} .(y) ^r .(2) ^r\end{gathered}
T
r+1
=
9
C
r
(x)
9−r
(2y)
r
T
r+1
=
9
C
r
(x)
9−r
.(y)
r
.(2)
r
Comparing with x⁶ y³ , we get, r = 3
Therefore,
\begin{gathered} \bf \: T _{r+1} \\ \: \tt ^9C_3 (x)^9−3 .y³ .2³ \\ \tt\: \: 9! (2)³× x⁶ × y³) / (3!.6!) \\ \: \tt 672x⁶ y³\end{gathered}
T
r+1
9
C
3
(x)
9
−3.y³.2³
9!(2)³×x⁶×y³)/(3!.6!)
672x⁶y³
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