# Question on normal distribution(again)

• September 15th 2010, 05:27 AM
mngeow
Question on normal distribution(again)
http://img835.imageshack.us/img835/4140/hardasscrap.jpg

I don't know how to do part (c) and (d)
• September 15th 2010, 05:39 AM
Quote:

Originally Posted by mngeow
http://img835.imageshack.us/img835/4140/hardasscrap.jpg

I don't know how to do part (c) and (d)

For (c), you are looking for the probability that $850, while eliminating the area of the graph up to $X=850$.

which is $\displaystyle\frac{P(X<1350)-P(X<850)}{P(X>850)}$

Convert to the Z-values and compute the probability.
Bear in mind that the probability of the bulb failing within the first 850 hours is now zero.
Hence your fraction has $P(X>850)$ in the denominator.

For (d), use the Binomial conversions $\sigma=\sqrt{npq},\;\;\; \mu=np$

where p=0.07, q=0.93, n=1200
• September 15th 2010, 05:41 AM
CaptainBlack
Quote:

Originally Posted by mngeow
http://img835.imageshack.us/img835/4140/hardasscrap.jpg

I don't know how to do part (c) and (d)

Let P(t) be the CDF of the bulb life.

For c you want [P(1350)-P(500)]/(1-P(500)) Imagin you have 1200 bulbs then 1200 P(500) have failed by 500 hours, 1200 P(1350) by 1350 hours, so out of the 1200 (1-P(500)) that survive the first 500 hours 1200 [P(1350)-P(500)] fail in the next 850 hours.

CB
• September 15th 2010, 05:59 AM
mngeow
ooh,I get it now,what about part d?
• September 15th 2010, 01:44 PM
CaptainBlack
Quote:

Originally Posted by mngeow
ooh,I get it now,what about part d?

The number that fail has a binomial distribution $B(0.07,1200)$, and you use the Normal approximation (and use a continuity correction) $N(0.07 \times 1200, 1200\times 0.07 \times 0.93)$

CB