# Thread: Normal Probability Distribution. Find X

1. ## Normal Probability Distribution. Find X

The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,225. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 795 pages. The manufacturer wants to provide guidelines to potential customers as to how long they can expect a cartridge to last.

How many pages should the manufacturer advertise for each cartridge if it wants to be correct 90 percent of the time? (Round z value to 2 decimal places. Round your answer to the nearest whole number.)

I tried z= 1.28, 1.29, didn't help. I consider it as 0.50 +0.40=0.90...What z should i use?

2. ## Re: Normal Probability Distribution. Find X

Standardize : Z=(X-μ)/σ=(Χ-12225)/795 ) and P(z)=P((X-μ)/σ)=0.900 .....check the tables...and find the answer

3. ## Re: Normal Probability Distribution. Find X

Shouls i check for P=0.40? I did everything like you wrote, but in my book the last value of Z is 3.09 and P = 0.4990

Z = 1.28 seems right, doesn't it?

4. ## Re: Normal Probability Distribution. Find X

Yes 1.28< z < 1.29 Φ(1.28)=0.8997 and Φ(1.29)= 0.9015
you may consider z = 1.28...

Sorry for this late reply ..I was out

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### the manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,200. the distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation

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