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The batteries from a certain manufacturer have a mean lifetime of 810 hours, with a standard deviation of 70 hours. Assuming that the lifetimes are normallydistributed, complete the following statements.(a) Approximately of the batteries have lifetimes between 670 hours and950 hours.(b) Approximately 68% of the batteries have lifetimes between 0 hours andhours

User Alextc
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PART A

Calculating the z score for 670 and 950

We have;


\begin{gathered} z_1=(670-810)/(70) \\ z_1=-(140)/(70) \\ z_1=-2 \\ \text{and } \\ z_2=(950-810)/(70) \\ z_2=(140)/(70)_{} \\ z_2=2 \end{gathered}


\begin{gathered} \text{from the graph } \\ \approx95\text{ \% of the batteries are within that range} \\ \end{gathered}

PART B

From the z score table we got z=0.47


\begin{gathered} 0.47=(x-810)/(70) \\ 70*\text{ 0.47 = X-810} \\ 32.9=X-810 \\ X=842.9 \end{gathered}

Approximately 68% of the batteries have lifetimes between 0 hours and

hours is 842.9



User Pbearne
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