- 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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- May 31st 2008, 06:20 PMcrashukcan some one help me please
- 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? - May 31st 2008, 07:08 PMcrashuk
correct its 1390-1400/300 = .033 then on the z table =.5120

p z >.033=.5120 - Jun 1st 2008, 01:08 AMmr fantastic
1. z = -0.033, NOT 0.033.

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. - Jun 1st 2008, 04:56 AMcrashuk
- 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 - Jun 1st 2008, 05:14 AMmr fantastic
- Jun 1st 2008, 07:32 AMcrashuk
- Jun 1st 2008, 02:09 PMmr fantastic
- Jun 1st 2008, 04:03 PMcrashuk