- The mean lifetime of a particular type of electric light bulb is 1400 hours, with a standard deviation of 300 hours.
What is the probability that this sample will yield a mean of less than 1390 hours?
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- The mean lifetime of a particular type of electric light bulb is 1400 hours, with a standard deviation of 300 hours.
What is the probability that this sample will yield a mean of less than 1390 hours?
correct its 1390-1400/300 = .033 then on the z table =.5120
p z >.033=.5120
1. z = -0.033, NOT 0.033.Quote:
Originally Posted by crashuk
2. You require Pr(Z < -0.033), NOT Pr(Z > 0.033) and not, for that matter, Pr(Z > -0.033).
3. Although I can figure out most of what you're saying in answer to your own question, if you wrote that way on a homework or test paper in a course of mine, you'd get points deducted from your grade. What you mean (I think) is:
"z = 1390-1400/300 = .033. Then from the z table, Pr(z < 0.033) =.5120"
(I have no idea where "p z >.033=.5120" has come from")
Sure, it takes a few more words to say that, but words matter. To communicate mathematics, you have to say exactly what you mean - no more, no less.
What is the probability that this sample will yield a mean of less than 1390 hours?
- The mean lifetime of a particular type of electric light bulb is 1400 hours, with a standard deviation of 300 hours.
so 1390-1400/300 = -0.033 = 51.20% probability that the sample would yield 1390 hours
which would = p(z < 0.033)=.5120
so if it a negative value then i have to look at negative values on the z table. which then would mean .4840
p(z < 0.033)=.4840
I get 0.4867.Quote:
Originally Posted by crashuk
Well, you can only be as accurate as the available tool - not your fault.Quote:
Originally Posted by crashuk
thanks mr fantastic youve been a help, sometimes getting the answer helps you understand.Quote:
Originally Posted by mr fantastic