jan_ai
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A construction company purchased concrete drill bits that have a lifespan that follows a
normal distribution with a mean of 75 hours and a standard deviation of 3 hours of use.
(i) Determine the percentage of the company’s drill bits that will have to be replaced
by or before 70 hours of use.
[3 marks]
(ii) Determine the percentage of the company’s drill bits that will have a lifespan
between 70 and 80 hours
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vc94
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(i) So the lifespan X has a normal distribution, X~N(75, 3^2). You want P(X <=70) .. can be done on calculator.
(ii) You want P(70 < X < 80), again can be done on the right calculator....
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jan_ai
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(Original post by vc94)
(i) So the lifespan X has a normal distribution, X~N(75, 3^2). You want P(X <=70) .. can be done on calculator.
(ii) You want P(70 < X < 80), again can be done on the right calculator....
how do you do this on a calculator
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vc94
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Does your calculator have statistical functions, like normal distribution?
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jan_ai
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(Original post by vc94)
Does your calculator have statistical functions, like normal distribution?
no
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vc94
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(Original post by jan_ai)
no
Have you learnt about standardising a normal distribution using Z= (X - mu)/sigma ?
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