## Angel_Kitty12 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?

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 $$\mu=k$$: $\begin{cases}\mu+\sigma =6500\\\mu+2\sigma=6800\end{cases}$