Question

Statistics help. Please
A company manufactures a brand of light bulb with a lifetimes in months that is normally distributed with mean 3 and variance 1. A consumer buys a number of these bulbs with the intention of replacing them successively as they burn out. The light bulbs have independent lifetimes.
What is the smallest number of bulbs to be purchased so that the succession of light bulbs produces light for at least 40 months with probability at least 0.9772?

Answer 1

\(X_1, X_2,\cdots, X_n\)~ \(N(3,1)\)

\(X=\sum_{i=1}^n X_i\)~ \(N(3n,n)\)

Answer 2

\(P(X \geq 40)\geq 0.97772\)

Answer 3

\(=1-P(X\leq 40)\geq 0.97772\)
Am I Ok up to now?

Answer 4

Yeah

Answer 5

okay..

Answer 6

\[1-P (\dfrac{X-its~mean}{its~standard~deviation}\leq \dfrac{40-its~mean}{its~standard~deviation})\geq 0.9772\]

Answer 7

That gives me.
\[P (\dfrac{X-its~mean}{its~standard~deviation}\leq \dfrac{40-its~mean}{its~standard~deviation})\leq 1- 0.9772=0.22228\]

Answer 8

right?

Answer 9

There is a calculation mistake in RHS

Answer 10

0.0228

Answer 11

where?

Answer 12

oh yes.

Answer 13

then, \(P(Z\leq \dfrac{40-3n}{\sqrt n})\leq 0.02228\)

Answer 14

Okay

Answer 15

use calculator, the invnorm, I have the ( ) =-2

Answer 16

Okays...

Answer 17

and solve for n, I got n = 16 and n =11.1

Answer 18

Perfect! I got 16

Answer 19

wait, let me double check,

Answer 20

(y) Take your time

Answer 21

Yes, but we have 2 roots, 16 and 11.1, why do you pick 16?

Answer 22

I actually used a different method... I got 16 and a negative root which I eliminated

Answer 23

P(T >= 40)
= \(\large{P(Z\ge \frac{40 - 3n}{\sqrt{n}})}\)
= \(\large{ 1 - \phi (\frac{40 - 3n}{\sqrt{n}})}\)
= \(\large{\phi(\frac{-40 + 3n}{\sqrt{n}}) }\)
= 0.9722

So, I got..

\(\large{\frac{-40 + 3n}{\sqrt{n}}}\) = 2

=> \(3n - 2\sqrt{n} - 40 = 0\)
=> \(\sqrt{n} = -10/3\) or 4

Answer 24

I ignored the other one.. and got n = 16

Answer 25

Also, n must be a whole number... we can't have fractional units of bulbs

Answer 26

The question is "at least", hence if we have fraction, then, take floor function. why not?
One more \(\sqrt n= -10/3\rightarrow n = 100/9\) . Either ways, you cannot get rid it off, right?

Answer 27

Hmm.. there is that too.. why don't we just plug in n = 100/9 and find out whether it satisfied our original condition or not

Answer 28

Yes, that is a good idea, Show me, please

Answer 29

One sec...

Answer 30

So, let's assume 12 bulbs

Answer 31

So, expected life = 36 months
variance = 12 => S.D. = sqrt(12)

Answer 32

Now, we need a lifetime of at least 40 months... i.e. 1.155 s.d. away from our expected life

Answer 33

What are the chances for that?

Answer 34

GGGGGGot you. thank you so much. I need completely understanding why I can get rid of the other root. Now, I know why. Again, thank you very much.

Answer 35

Glad I could help.. it's been a whole year since I last read probability distributions :P

Answer 36

btw, I got solution for my PDE problem also :)