# Urgent quick stats help please!

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Thread starter 4 years ago
#1
I wasnt in lesson today as i had a trip and my teacher sent me a sheet to pratice. I dont remember how to do the following and ive left my folder in school so cannot look at my notes. Any help is appeciated, thanks in advance.

The random variable X is normally distributed with a mean of 42 and a variance of 18. Find

a) P( X </= 45)

b) P(32 </= X </= 38)

Im guessing for a) you do P[X</= (45-42)/18] if that makes any sense, sorry
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4 years ago
#2
(Original post by Aty100)
I wasnt in lesson today as i had a trip and my teacher sent me a sheet to pratice. I dont remember how to do the following and ive left my folder in school so cannot look at my notes. Any help is appeciated, thanks in advance.

The random variable X is normally distributed with a mean of 42 and a variance of 18. Find

a) P( X </= 45)

b) P(32 </= X </= 38)

Im guessing for a) you do P[X</= (45-42)/18] if that makes any sense, sorry
That is a good start

Notation wise, you go from P(X<= x) to P(Z <= (x-mu)/sd) where mu is the mean of X and sd is the standard deviation of X. You have to be careful as you've been given variance, not standard deviation.

Once you've calculated what (x-mu)/sd is, you can appeal to the tables of z values (which usually give P(Z<=z)) and in Q1 you should be able to directly read it off.

Edit: it should've been x rather than an a from nowhere.
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Thread starter 4 years ago
#3
(Original post by SeanFM)
That is a good start

Notation wise, you go from P(X<= a) to P(Z <= (x-mu)/sd) where mu is the mean of X and sd is the standard deviation of X. You have to be careful as you've been given variance, not standard deviation.

Once you've calculated what (x-mu)/sd is, you can appeal to the tables of z values (which usually give P(Z<=z)) and in Q1 you should be able to directly read it off.
Thanks so much,
So the SD would be root18 ?

So i get z as 0.707106.. Would i look for 0.70 or 0.71 in the table ?
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4 years ago
#4
(Original post by Aty100)
Thanks so much,
So the SD would be root18 ?
Correct.
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Thread starter 4 years ago
#5
(Original post by SeanFM)
Correct.
Yay so
(45-42)/root18
= 0.707106...

So if im correct
(Z<=0.71) ? Or so i use 0.70?
Then i look at my table and the answer is the value i see

For 0.70 it would be 0.7580
For 0.71 it would be 0.7611
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4 years ago
#6
(Original post by Aty100)
Yay so
(45-42)/root18
= 0.707106...

So if im correct
(Z<=0.71) ? Or so i use 0.70?
Then i look at my table and the answer is the value i see

For 0.70 it would be 0.7580
For 0.71 it would be 0.7611
Yes, I'd suggest rounding to the nearest 2 d.p.
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Thread starter 4 years ago
#7
(Original post by SeanFM)
Yes, I'd suggest rounding to the nearest 2 d.p.
So for part b) if you don't mind checking:

P[(32-42)/root18 <= Z <= (38-42)/root18]

So

= P(-2.357 <= Z <= -0.942)

= P(Z </= -0.942) - P(Z </= -2.357)

= 1-P(Z >/= 0.942)
= 1-0.864
=0.136 ?

Then same method for -2.357

Then minus the figure I get from 0.136 to get my answer ?
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4 years ago
#8
(Original post by Aty100)
So for part b) if you don't mind checking:

P[(32-42)/root18 <= Z <= (38-42)/root18]

So

= P(-2.357 <= Z <= -0.942)

= P(Z </= -0.942) - P(Z </= -2.357)

= 1-P(Z >/= 0.942)
= 1-0.864
=0.136 ?

Then same method for -2.357

Then minus the figure I get from 0.136 to get my answer ?
Note that P(Z<z) where z is a negative number is equal to P(Z>-z) using symmetry. So P(Z </= -0.942) = P(Z => 0.942) = 1-P(Z<=0.942). (which is what we can read off the tables).

Anyway, you appear to have not found P(Z<=-0.94) correctly. I think you've somehow got the probability for P(Z<=1.10) in there (which is 0.864) but your calculations are correct.
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Thread starter 4 years ago
#9
(Original post by SeanFM)
Note that P(Z<z) where z is a negative number is equal to P(Z>-z) using symmetry. So P(Z </= -0.942) = P(Z => 0.942) = 1-P(Z<=0.942). (which is what we can read off the tables).

Anyway, you appear to have not found P(Z<=-0.94) correctly. I think you've somehow got the probability for P(Z<=1.10) in there (which is 0.864) but your calculations are correct.
Thanks so much for your help. Yah sorry it should have been 0.8264
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Thread starter 4 years ago
#10
(Original post by SeanFM)
Note that P(Z<z) where z is a negative number is equal to P(Z>-z) using symmetry. So P(Z </= -0.942) = P(Z => 0.942) = 1-P(Z<=0.942). (which is what we can read off the tables).

Anyway, you appear to have not found P(Z<=-0.94) correctly. I think you've somehow got the probability for P(Z<=1.10) in there (which is 0.864) but your calculations are correct.
So far the last bit I would do
1-0.8264
=0.1736

And then do what I said before?
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4 years ago
#11
(Original post by Aty100)
So far the last bit I would do
1-0.8264
=0.1736

And then do what I said before?
Correct well done
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Thread starter 4 years ago
#12
So then I do

= 1-P(Z >/= 2.357)
= 1-0.9909
=0.0091

So finally I complete the question by doing:

0.1736 - 0.0091
= 0.1645 ?
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4 years ago
#13
(Original post by Aty100)
So then I do

= 1-P(Z >/= 2.357)
= 1-0.9909
=0.0091

So finally I complete the question by doing:

0.1736 - 0.0091
= 0.1645 ?
I have to be pernickety and say that it's 1 - P(Z less than or equal to 2.357) rather than 1-P(Z greater than or equal to 2.357) but you have looked up the right value and got the correct answer anyway.

(Remember that the tables give P(Z less than or equal to z).
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Thread starter 4 years ago
#14
(Original post by SeanFM)
I have to be pernickety and say that it's 1 - P(Z less than or equal to 2.357) rather than 1-P(Z greater than or equal to 2.357) but you have looked up the right value and got the correct answer anyway.

(Remember that the tables give P(Z less than or equal to z).
Yah thank you I'm just tried and making silly mistakes. By I'm so grateful for how much you help me!
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4 years ago
#15
(Original post by Aty100)
Yah thank you I'm just tried and making silly mistakes. By I'm so grateful for how much you help me!
No problem. You did a good job without notes.
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Thread starter 4 years ago
#16
(Original post by SeanFM)
I have to be pernickety and say that it's 1 - P(Z less than or equal to 2.357) rather than 1-P(Z greater than or equal to 2.357) but you have looked up the right value and got the correct answer anyway.

(Remember that the tables give P(Z less than or equal to z).
How would I work out the interquartile range?
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4 years ago
#17
(Original post by Aty100)
How would I work out the interquartile range?
Well, what is the definition of interquartile range?

And so how would you find it?
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Thread starter 4 years ago
#18
(Original post by SeanFM)
No problem. You did a good job without notes.
It is getting quite late so if you can't help me tonight it's fine. Just needed help with interquartile range because that's what my class were doing today whilst I was away and I just was some knowledge and understanding of it for when we do it on Friday.
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Thread starter 4 years ago
#19
(Original post by SeanFM)
Well, what is the definition of interquartile range?

And so how would you find it?
IQR = Q3 - Q1 ?
That's pretty much all I remember from GCSEs maths I didn't do stats.
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4 years ago
#20
(Original post by Aty100)
It is getting quite late so if you can't help me tonight it's fine. Just needed help with interquartile range because that's what my class were doing today whilst I was away and I just was some knowledge and understanding of it for when we do it on Friday.
See the previous question that I posed to you

What's the definition of interquartile range? (You may have seen it in GCSE stats
if you have done it or Maths).
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