• Angel_Kitty12
Can someone please help me? The Normal Distribution (Algebra 2) LIGHT BULBS: The time that a certain brand of light bulb last before burning out is normally distributed. About 2.5% of the bulbs last longer than 6800 hours and about 16% of the bulbs last longer than 6500 hours. How long does the average bulb last?
  • Stacey Warren - Expert
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  • jamiebookeater
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  • anonymous
You have that \(P(X>6800)=0.025=2.5\%\) and \(P(X>6500)=0.16=16\%\), and you want to find \(k\) such that \(P(X>k)=0.5\). This is because the normal distribution is symmetric about its mean, i.e. if \(\mu\) is the mean of a normally distributed random variable \(X\), then \(P(X>\mu)=P(X<\mu)=0.5=50\%\). Recall the empirical rule: \(68\%\) of the distribution lies within one standard deviation of the mean, \(95\%\) within two standard deviations, and \(99.7%\) within three standard deviations. The last part tells you that \(0.3\%\) of the distribution lies outside \[ \begin{matrix} \begin{cases}P(X>6800)=1-P(X<6800)&(\clubsuit)\\ P(X>6500)=1-P(X<6500)&(\spadesuit)\end{cases}\\[1ex] \Downarrow\\[1ex] -(\clubsuit)+(\spadesuit)\\[1ex] \Downarrow\\[1ex] \begin{align*}-\bigg(1-P(X<6800)\bigg)+\bigg(1-P(X<6500)\bigg)&=P(X<6800)-P(X<6500)\\[1ex] -0.025+0.16&=P(6500

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