Need help on this s2 concept please asap

Original post by optic123

S2 ANYONE, I DONT GET THIS CONCEPT. ITS HURTING MY HEAD SO MUCH. PLEASE CAN SOMEONE HELP.

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The length is an estimate as it's rounded to the nearest cm (by +/- 0.5) so your interval would be [19.5, 20.5] (I think) as it's a uniform distribution ( the real length could be anywhere between that interval).

For b. E(X) is the midpoint and for Var(x) you can use the formula

(edited 9 years ago)

Reply 2

OP

Original post by kkboyk

The length is an estimate as it's rounded to the nearest cm (by +/- 0.5) so your interval would be [19.5, 20.5] (I think) as it's a uniform distribution ( the real length could be anywhere between that interval).

For b. E(X) is the midpoint and for Var(x) you can use the formula

dont get what you mean? sorry?

Reply 3

Original post by optic123

dont get what you mean? sorry?

For what part in particular?

Reply 4

OP

Original post by kkboyk

For what part in particular?

all of it, could you possible do the working out and then explain using that ?

Reply 5

Original post by optic123

all of it, could you possible do the working out and then explain using that ?

Actually ignore what I've said earlier. I can't do the working out since I don't have paper with me.

For part a: the child had cut the string on the point marked with x, and x can be anywhere between 0<x<20 so the interval would be [0,20]. It's a continuous distribution because x can be anywhere between between that inequality.

for b: The mean is just the midpoint of the interval [0, 20]. The variance you have to use the formula (b^2 - a^2)/12 (you have to check the formula book, I think I'm wrong)

Reply 6

OP

Original post by kkboyk

Actually ignore what I've said earlier. I can't do the working out since I don't have paper with me.

For part a: the child had cut the string on the point marked with x, and x can be anywhere between 0<x<20 so the interval would be [0,20]. It's a continuous distribution because x can be anywhere between between that inequality.

for b: The mean is just the midpoint of the interval [0, 20]. The variance you have to use the formula (b^2 - a^2)/12 (you have to check the formula book, I think I'm wrong)

For part c ) thats where it get tricky, detail would be great. Dumb it down as much as possible.

Reply 7

Original post by optic123

For part c ) thats where it get tricky, detail would be great. Dumb it down as much as possible.

Find the probability that X is bigger than 8, so now you're trying to find the area between [8,20].

Reply 8

OP

Original post by kkboyk

Find the probability that X is bigger than 8, so now you're trying to find the area between [8,20].

That is incorrect.

Reply 9

Dalek1099

Original post by optic123

That is incorrect.

You want the probability the length of string is between 8 and 12 as either piece of string could be the shorter piece,if the piece is longer than 12 then the other piece is the shorter one and less than 8.

Reply 10