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  • one year ago

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?

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  1. anonymous
    • one year ago
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    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 \[ \begin{matrix} \begin{cases}P(X>6800)=1-P(X<6800)&(\clubsuit)\\ P(X>6500)=1-P(X<6500)&(\spadesuit)\end{cases}\\[1ex] \Downarrow\\[1ex] -(\clubsuit)+(\spadesuit)\\[1ex] \Downarrow\\[1ex] \begin{align*}-\bigg(1-P(X<6800)\bigg)+\bigg(1-P(X<6500)\bigg)&=P(X<6800)-P(X<6500)\\[1ex] -0.025+0.16&=P(6500<X<6800)\\[1ex] 0.135&=P(6500<X<6800) \end{align*} \end{matrix}\] This happens to be exactly the probability that \(X\) falls between one and two standard deviations from the mean. In other words, if \(\mu\) is the mean and \(\sigma\) is the standard deviation, you have \[P(\mu+\sigma<X<\mu+2\sigma)=0.135\] This means you have the system with two unknowns, where you're interested in finding \(\mu=k\): \[\begin{cases}\mu+\sigma =6500\\\mu+2\sigma=6800\end{cases}\]

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