Normal distribution

I want you to check my calculations in the following question:

A manufacturer makes two sizes of elastic bands: large and small. 40% of the bands produced are large bands and 60% are small bands. Assuming that each pack of these elastic bands contains a random selection, calculate the probability that, in a pack containing 20 bands, there are
(i) equal number of large and small bands.

20C10 (0.40)^10(0.60)^10 = 0.117

(ii) more than 17 small bands.

Here I appled normal ditsribution because np = 20 * 0.60 =12 and nq= 8 ..both of them are greater than 8 so binomial can not be applied ..here is what i get when i apply normal distribution
P(Z > 17.5 - 12 / sqrt 4.8) = P(Z >2.510) = 1- 0.9940 = .006

but what I don't understand is that even in part(a) np and nq are greater than 5 but I still use binomial ...is that right ..i can't figure out another way to solve part(a)

(iii)An office pack contains 150 elastic bands.
Using a suitable approximation, calculate the probability that the number of small bands in the office pack is between 88 and 97.

My calculation :

X ~ N(90, 36)

P(87.5 <= X <= 97.5) = P ( -0.417 <=Z <= 1.25)
= 0.8944 - 1 + 0.6616 = 0.556

Reply 1

73337

18

coursework.info is right. again.

That guy is a machine.

Reply 2

DFranklin

18

For part (ii) I'd still use binomial. Just add the probabilities for 18,19 and 20 small bands.

Reply 3

e-n-i-g-m-a

OP

DFranklin

For part (ii) I'd still use binomial. Just add the probabilities for 18,19 and 20 small bands.

why would you use binomial ...isn't nq and np greater than 5 ???

Reply 4

DFranklin

18

e-n-i-g-m-a

why would you use binomial ...isn't nq and np greater than 5 ???

Who cares? It's a binomial distribution, and it is not much work to do the exact calculation, so why would you use a normal approximation?

Reply 5

e-n-i-g-m-a

OP

but you get a really small answer when you use binomial !!

Reply 6

DFranklin

18

e-n-i-g-m-a

but you get a really small answer when you use binomial !!

It will be the right answer, though. Treating the distribution as normal is only an approximation.

Reply 7

e-n-i-g-m-a

OP

I still don't understand the nq and np thing !!!

Reply 8

insparato

15

e-n-i-g-m-a

I still don't understand the nq and np thing !!!

nq, np?

Reply 9

jakezg

e-n-i-g-m-a

I still don't understand the nq and np thing !!!

lol you are right to use the normal approximation of X~B(n, p) when n is large, np>5 and np(1-p)>5.

This becomes Y~n(np, np(1-p)).. be careful with the np(1-p) though as its the variance and not the SD. Remeber continuity corrections.

Reply 10

DFranklin

18

jakezg

lol you are right to use the normal approximation of X~B(n, p) when n is large, np>5 and np(1-p)>5.Sigh.

Whatever the values of np, n(1-p), the normal approximation is only that, an approximation. It is reasonable to argue that because np, n(1-p) are > 5, it is a sufficiently good approximation. But to argue that because np,n(1-p) are > 5 it is actually correct and the binomial is not is not at all reasonable.

In this case, it takes very little effort to directly calculate the probablity of more than 17 elastic bands, and as such, I would personally say the use of a normal approximation is not appropriate. Particularly since the relative error involved in using a normal approximation is much higher near the tail ends of the distribution.

Note that the question makes no mention of approximation for part (ii); it is only for part (iii) that it says to use a suitable approximation. So I would think the question setter agrees with me.

Reply 11

em1n3m

e-n-i-g-m-a

I want you to check my calculations in the following question:

A manufacturer makes two sizes of elastic bands: large and small. 40% of the bands produced are large bands and 60% are small bands. Assuming that each pack of these elastic bands contains a random selection, calculate the probability that, in a pack containing 20 bands, there are
(i) equal number of large and small bands.

20C10 (0.40)^10(0.60)^10 = 0.117

(ii) more than 17 small bands.

Here I appled normal ditsribution because np = 20 * 0.60 =12 and nq= 8 ..both of them are greater than 8 so binomial can not be applied ..here is what i get when i apply normal distribution
P(Z > 17.5 - 12 / sqrt 4.8) = P(Z >2.510) = 1- 0.9940 = .006

but what I don't understand is that even in part(a) np and nq are greater than 5 but I still use binomial ...is that right ..i can't figure out another way to solve part(a)

(iii)An office pack contains 150 elastic bands.
Using a suitable approximation, calculate the probability that the number of small bands in the office pack is between 88 and 97.

My calculation :

X ~ N(90, 36)

P(87.5 <= X <= 97.5) = P ( -0.417 <=Z <= 1.25)
= 0.8944 - 1 + 0.6616 = 0.556

LOOOL... Man, it's like ****ing Deja-Vu. This is from the Nov06 CIE paper 6 isn't it? This is the exam I sat, and I stuffed this question up, although it's so easy :mad:

Still got A though :biggrin:

I'll look at your working, give me a sec

Reply 12

em1n3m

Ok your answers and methods are right... For the second part (ii), you can use normal or binomial; they're both right... If you want the mark scheme just PM me, it shows both.

Normal distribution

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