Solution
Let lifetime of bulbs be x
X~N"(1150, {175} ^2)"
Pr(x>1000)=1-Pr(x<1000)
Obtaining the z-score of the above yields :
"\\Phi ({1000-1150 \\over 175}) = \\Phi (-0.85714)"
Which follows x~N(0,1)
We can thus obtain the probability from statical tables.
"\\Phi (-0.85714) =0.1957"
"1-\\Phi (-0.85714)=0.8043"
Ans : 0.8043
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