A company tested 735 of the lightbulbs they produced and found them to have a mean life of 1,200 hours and a standard deviation of 50 hours. How many of these lightbulbs had a life between 1,170 hours and 1,230 hours?
P(1170≤X≤1230)=P(1170−120050≤Z≤1230−120050)=F(1230−120050)−F(1170−120050)=F(0.6)−F(−0.6)=0.7257−0.2743=0.4514P(1170\le X \le1230)=\\ P(\frac{1170-1200}{50}\le Z \le\frac{1230-1200}{50})=\\ F(\frac{1230-1200}{50})-F(\frac{1170-1200}{50})=\\ F(0.6)-F(-0.6)=\\ 0.7257 -0.2743 =\\ 0.4514P(1170≤X≤1230)=P(501170−1200≤Z≤501230−1200)=F(501230−1200)−F(501170−1200)=F(0.6)−F(−0.6)=0.7257−0.2743=0.4514
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