SOLUTION: A certain type of thermal bettery for an airplane navigation device backup power has a mean life of 300 hours with a standard deviation of 15 hours. What proportion of these batter
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Question 280196: A certain type of thermal bettery for an airplane navigation device backup power has a mean life of 300 hours with a standard deviation of 15 hours. What proportion of these batteries can be expected to have lives of 322 hours or less? Assume a normal distribution of backup power devices lives.
I'm so confused about what I am supposed to do. Here is what I have tried:
15+300/322= 5130/322=15.9
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A certain type of thermal bettery for an airplane navigation device backup power has a mean life of 300 hours with a standard deviation of 15 hours. What proportion of these batteries can be expected to have lives of 322 hours or less? Assume a normal distribution of backup power devices lives.
I'm so confused about what I am supposed to do. Here is what I have tried:
-------------------
To get unconfused:
Sketch a normal curve.
Put 300 at the middle of the x-axis under the curve.
Put 15 under the 300 just to remind yourself that 15 is the std.
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Mark 322 at an apporiate point on the x-axis.
Shade the area under the curve and to the left of 322.
That shaded area is the answer to your problem.
-------------------------
Find the z-value of 322:
z(322)= (322-300)/15 = 1.467
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The following you just have to understand.
Use your z-chart or your calculator to get:
P(x < 322) = P(z < 1.467) = 0.9288
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Cheers,
Stan H.
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