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QUESTION
 The length of a certain type of battery is normally distributed
  with mean 5.0cm and standard deviation .05cm. Find the
  probability that such a battery has a length between 4.92 and
  5.08 cm.\\
  tubes are manufactured to contain 4 such batteries. 95\% of the
  tubes have lengths greater than 20.9 and 10\% have lengths
  greater than 21.6cm. Assuming that the lengths of the tubes are
  also normally distributed, find the mean and standard deviation
  of the lengths correct to two decimal places.\\
  If tubes and batteries are chosen independently, find the
  probability that a tube will contain 4 batteries with at least
  0.75cm to spare.

ANSWER
 $B\sim N(5.0,0.5^2)$\\

  \begin{eqnarray*}
   P(4.92<B<5.08)&=&\Phi(\frac{5.08-5.0}{0.05})-\Phi(\frac{4.92-5.0}{.05})\\
   &=&\Phi(1.6)-\Phi(-1.6)=0.9452-(1-0.9452)\approx 0.89
  \end{eqnarray*}

  $95\%>20.9\rightarrow 20.9 =\mu-1.6449\sigma\\
  10\%>21.6\rightarrow 21.6=\mu+1.2816\sigma$\\
  Subtracting the two equations gives $0.7=2.9265\sigma,\
  \sigma=0.2392=0.24$ to 2 d.p.Substituing in either equation gives
  that $\mu=21.29$ to 2 d.p.\\
  Distribution of $B_1+B_2+B_3+B_4 \sim N(20.0,4\times
  0.05^2)=N(20.0,0.01)\\
  T-(B_1+B_2+B_3+B_4) \sim N(1.29,0.24^2+0.01)$\\

  \begin{eqnarray*}
   P(T-(B_1+B_2+B_3+B_4)>0.75)&=&1-\Phi(\frac{0.75-1.29}{sqrt{0.0676}})\\
   &=&1-\Phi(-\frac{0.54}{0.26})\\
   &=&1-\Phi(-2.077)=\Phi(2.077)\\
   &=&0.981
  \end{eqnarray*}


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