# Urgent quick stats help please!

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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

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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#2

(Original post by

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

**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

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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(Original post by

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.

**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.

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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(Original post by

Correct.

**SeanFM**)Correct.

(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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#6

(Original post by

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

**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

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(Original post by

Yes, I'd suggest rounding to the nearest 2 d.p.

**SeanFM**)Yes, I'd suggest rounding to the nearest 2 d.p.

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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#8

(Original post by

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 ?

**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 ?

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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(Original post by

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.

**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.

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**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.

1-0.8264

=0.1736

And then do what I said before?

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#11

(Original post by

So far the last bit I would do

1-0.8264

=0.1736

And then do what I said before?

**Aty100**)So far the last bit I would do

1-0.8264

=0.1736

And then do what I said before?

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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 ?

= 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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#13

(Original post by

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 ?

**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 ?

(Remember that the tables give P(Z less than or equal to z).

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(Original post by

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).

**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).

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#15

(Original post by

Yah thank you I'm just tried and making silly mistakes. By I'm so grateful for how much you help me!

**Aty100**)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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**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).

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#17

(Original post by

How would I work out the interquartile range?

**Aty100**)How would I work out the interquartile range?

And so how would you find it?

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(Original post by

No problem. You did a good job without notes.

**SeanFM**)No problem. You did a good job without notes.

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(Original post by

Well, what is the definition of interquartile range?

And so how would you find it?

**SeanFM**)Well, what is the definition of interquartile range?

And so how would you find it?

That's pretty much all I remember from GCSEs maths I didn't do stats.

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#20

(Original post by

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.

**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.

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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