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S1- Normal Distribution

Can someone help me with part b and c? Not sure on what do do.
Thanks in advance :smile:
S1 - Normal distribution.jpg42.9KB
Original post by Super199
Can someone help me with part b and c? Not sure on what do do.
Thanks in advance :smile:


So you know that

TT~N(25,42)N(25, 4^2)

Let T−25=Y⇒YT - 25 = Y \Rightarrow Y~N(0,42)N(0, 4^2)

Now P(∣T−25∣<5)⇒P(∣Y∣<5)P(|T-25| < 5) \Rightarrow P(|Y| < 5)

Do you know what the modulus sign represents in the normal distribution?

P(∣Y∣<5)P(|Y| < 5) is just the same as P(−5<Y<5)P(-5 < Y < 5)
(edited 9 years ago)
Reply 2
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Super199
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Original post by CTArsenal
So you know that

T N(25,42)T~N(25, 4^2)

Let T−25=YT - 25 = Y ⇒Y N(0,42)\Rightarrow Y~N(0, 4^2)

Now P(∣T−25∣<5)⇒P(∣Y∣<5)P(|T-25| < 5) \Rightarrow P(|Y| < 5)

Do you know what the modulus sign represents in the normal distribution?

P(∣Y∣<5)P(|Y| < 5) is just the same as P(−5<Y<5)P(-5 < Y < 5)

oh wow I have never been taught that. Will give it another go, cheers :smile:
Edit: I'm sort of struggling to understand how |Y|<5 gives P(-5<Y<5)?
(edited 9 years ago)
Original post by Super199
oh wow I have never been taught that. Will give it another go, cheers :smile:


Haha don't worry, I'm doing S4 and started to understand it from there! Never learnt it before that oddly enough.

For part (c), all you do is [P(T<23)]3[P(T < 23)]^3 :smile:
Original post by Super199
oh wow I have never been taught that. Will give it another go, cheers :smile:
Edit: I'm sort of struggling to understand how |Y|<5 gives P(-5<Y<5)?


If you think of it in terms of integers then
Unparseable latex formula:

|Y|<5 \text { implies } Y \text{ can take the values } -4,-3,-2,-1,0,1,2,3 & 4


More generally, since we define ∣Y∣=−Y when Y<0 then this gives −Y<5⇒Y>−5|Y|=-Y \text{ when }Y<0 \text { then this gives }-Y<5 \Rightarrow Y>-5
 and of course if Y>0 then Y<5 \text{ and of course if }Y>0 \text{ then } Y<5
Putting them together −5<Y<5-5<Y<5
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Original post by CTArsenal
Haha don't worry, I'm doing S4 and started to understand it from there! Never learnt it before that oddly enough.

For part (c), all you do is [P(T<23)]3[P(T < 23)]^3 :smile:

For part c the mark scheme says P(T<23)^3=P(Z<0.5)^3
Why isn't it P(Z<-0.5)?
23-25/4 = -0.5?
Original post by Super199
For part c the mark scheme says P(T<23)^3=P(Z<0.5)^3
Why isn't it P(Z<-0.5)?
23-25/4 = -0.5?


Looks like there's an error in the mark scheme, see if the answer you get is the same as them with [P(Z<−0.5)]3[P(Z < -0.5)]^3.
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Original post by CTArsenal
Looks like there's an error in the mark scheme, see if the answer you get is the same as them with [P(Z<−0.5)]3[P(Z < -0.5)]^3.

They get 0.3307
and I get 0.02937 .. haha :rolleyes:
Original post by Super199
They get 0.3307
and I get 0.02937 .. haha :rolleyes:


Definitely an error in the paper's mark scheme then, especially if they'd already stated [P(T<23)]3[P(T < 23)]^3 as a basis for the calculation.
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Original post by CTArsenal
Definitely an error in the paper's mark scheme then, especially if they'd already stated [P(T<23)]3[P(T < 23)]^3 as a basis for the calculation.

Cool, thanks for your help :smile:
Reply 10
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Homies I need some help

Given that the mean =27 and sigma^2=10

b) Find P(26<x<28)

Is the standard deviation 10 or square root of 10?

I got 0.248 as my final answer can someone check that for me.

Safe
Original post by Super199
Homies I need some help

Given that the mean =27 and sigma^2=10

b) Find P(26<x<28)

Is the standard deviation 10 or square root of 10?

I got 0.248 as my final answer can someone check that for me.

Safe


Not sure if I'm right but 10 is the standard deviation and then used 28-27/10 to get 0.1 on tables this is 0.5398 then I did 0.5398-0.5 to get the probability of one side which is 0.0398 then times it by 2 because of the symmetry as they would have the same probability because they're one space apart then got 0.0796
Reply 12
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Original post by studentwiz
Not sure if I'm right but 10 is the standard deviation and then used 28-27/10 to get 0.1 on tables this is 0.5398 then I did 0.5398-0.5 to get the probability of one side which is 0.0398 then times it by 2 because of the symmetry as they would have the same probability because they're one space apart then got 0.0796

I've done it now but cheers anyway
Original post by Super199
I've done it now but cheers anyway


was it right?
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Original post by studentwiz
was it right?

Yeah

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