Further probability question

eoins23456

Bit unsure about this question n need someone to verify the solutions please:D

Q1-A company sells 20000 computers per year.The mean life expectancy of these computers is ten years with a standard deviation of 1.8 years.

Calculate the probability that a randomly chosen computer will breakdown before it is eight years old?

Just popped everything into the formula z=x-mean/deviation

8-10/1.8 giving me 1.11 =1-.8665=.1335=probability(page 36 of the tables)

what is the probability that a computer will still be working after 15 years?

did the same thing 15-10/1.8=2.78=1-.9973=.0027=probability

Bit unsure of my solution as i didnt use the 20000 computers bit.



A car manufacturer claims thats it cars consume on average 11.5 litres of petrol for each 100km driven.N of these car are tested and it was found that the average consumption was 11.9 litres for each 100km with a standard deviation of 2.1 litres. find the greatest value of N for which the manufacturers claim would not be rejected at the five percent level of significance.

square root of N(11.9-11.4)/2.1=1.96

works out at n equal to 67

Can anyone verify or correct these solutions?

leavingcert

for the first question the 20000 comouters equals n

You never got the S.E. the formula you shouldve used was X-mean over the S.E which is standard deviation over square root of n

Now, im wondering if you could help me with something,

I did this question by myself coz i couldnt understand question 8 and love probability im just wondering you know in the question similar to your first (a) section

when you get say -1.3 >z>1.2

how do you know when to minus the number from the book of tables from one or not?

eoins23456

i tried using that method first but got like 200 for an answer which doesnt make sense bit lost with the question.

-1.3<z<1.2
In general u get the probability of the one one on the right p(z<1.2) first then minus the one on the left if the one of the left is a minus number change the sign of the inequality so it would be 1 minus p(Z>1.3)
answer=p(z<1.2)-(1-p(z>1.3))

If its 1<Z<2
Just get p(Z<2)-P(z<1)

eoins23456

Trying to teach myself the option as well cos it can be done quicker then question 8 IMO.

leavingcert

Ahhhh!
I get it!

Thanks so much i was just chancin it everytime i did it :P

I konw1 and it is so much easier isnt it?
And the answers are like an A4 page in length whenm worked out fully and in the marking scheme you get 5 marks for getting the meand and another 5 marks for the deviation.

Im just worried because if you dont get the question for some reason like you read it wrong its 20m gone, no attempt marks if the figures are wrong.

But let's hope it works out for us in 2 weeks time :P

eoins23456

yeah i know there handing out marks for nothing.U could even get the question done in 10-15 mins if its a nice one.just try a few examples and u shud be on your way.I was chancing my arm for like ages before i finally got the hang of it:P lets just hope for an easy question haha

MathsManiac

Getting back to your original questions:

1st one is correct, other than the fact that you left out a minus sign in front of the 1.11. The 20000 is not relevant The question doesn't state that the life expectancy is normally distributed, but if this is intended to be an LC question, then that's the only distribution it could be. (Q. should have said it, though.)

Last one is also correct. (Strictly speaking, a t-test might be more appropriate than a z-test here, but that's not on the LC course, so a z-test is fine. Anyway, they're pretty close to eachother for n>30.)