# Thread: Question on normal distribution(again)

1. ## Question on normal distribution(again)

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

2. Originally Posted by mngeow

I don't know how to do part (c) and (d)
For (c), you are looking for the probability that $\displaystyle 850<X<1350$, while eliminating the area of the graph up to $\displaystyle X=850$.

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

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 $\displaystyle P(X>850)$ in the denominator.

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

where p=0.07, q=0.93, n=1200

3. Originally Posted by mngeow

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

4. ooh,I get it now,what about part d?

5. Originally Posted by mngeow
ooh,I get it now,what about part d?
The number that fail has a binomial distribution$\displaystyle B(0.07,1200)$, and you use the Normal approximation (and use a continuity correction) $\displaystyle N(0.07 \times 1200, 1200\times 0.07 \times 0.93)$

CB