Please can someone help me with the following questions. We only started learning S1 the week before half term holidays and I was absent for a few lessons. (was ill on mon and tues and that when we have 2 lessons of stats per day) so I don't really understand the tasks ive been set as ive never done stats before . Any help will be much appreciated. Thanks in advance.
Consider the distribution X~N(16,9) to calculate the following probability:
1) P(10<X<12)
Consider the distribution X~N(85,25) to calculate the following probability:
1) P(80<X<100)
2) The value of a if P(X>a) = 0.67
Consider the distribution X~N(15,16) to calculate the following probability:
1) P(16<X<19)
2) The value of a if P(X<a) = 0.975

Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 1
 30102015 17:21

Kevin De Bruyne
 Follow
 609 followers
 21 badges
 Send a private message to Kevin De Bruyne
 Very Important Poster
Offline21ReputationRep:Very Important Poster Follow
 2
 30102015 17:26
(Original post by Aty100)
x
This page seems to have a few videos on the normal distribution.Last edited by Kevin De Bruyne; 30102015 at 17:29. 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 3
 30102015 17:29
(Original post by SeanFM)
I would suggest getting the textbook, reading through them and attempting the questions and post here with your working if you're stuck 
Kevin De Bruyne
 Follow
 609 followers
 21 badges
 Send a private message to Kevin De Bruyne
 Very Important Poster
Offline21ReputationRep:Very Important Poster Follow
 4
 30102015 17:31
(Original post by Aty100)
I don't have a textbook
See the link in the above post for some videos, or find some if that's not a useful one. 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 5
 30102015 17:33
(Original post by SeanFM)
Get a hold of one if you can, they are really useful.
See the link in the above post for some videos, or find some if that's not a useful one. 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 6
 30102015 17:59
(Original post by SeanFM)
Get a hold of one if you can, they are really useful.
See the link in the above post for some videos, or find some if that's not a useful one.
= P(1016/3 < Z < 1216/3)
= P(2 < Z < 1.33)
now do I do
= 1 P(Z > 1.33)  1  P(Z > 2)
??
I'm stuckLast edited by Aty100; 30102015 at 18:01. 
Kevin De Bruyne
 Follow
 609 followers
 21 badges
 Send a private message to Kevin De Bruyne
 Very Important Poster
Offline21ReputationRep:Very Important Poster Follow
 7
 30102015 18:16
(Original post by Aty100)
so for the first one I did
= P(1016/3 < Z < 1216/3)
= P(2 < Z < 1.33)
now do I do
= 1 P(Z > 1.33)  1  P(Z > 2)
??
I'm stuck
So you have P(2 < Z < 1.33), which can be written as P(Z<1.33)  P(Z<2). Why? You may wish to use a number line to help you.
You can split the range Z < 1.33 into two different regions: 2 < Z < 1.33 (which is what we want) and the rest is Z<2, and combining those two regions together again gives Z <  1.33. So if we express P(2 < Z < 1.33) in terms of P(Z <  1.33) then we would be counting the area 'Z < 2' but we're not interested in that, only the region 2 < Z < 1.33, so that's why it can be written as P(Z<1.33)  P(Z<2). That's just for your information/understanding.
Now the problem is that the tables have values P(Z<z) where z is a positive number, and you have to use properties of the normal distribution when you're dealing with things like P(Z>z) or P(Z<z) (where z is a positive number) or even P(Z>z).
So we've got P(Z<1.33)  P(Z<2) to work out. We need everything to be of the form P(Z<z), where z is a positive number.
So you use two things (one of which you've done already). P(Z<z) = P(Z>z) by symmetry. This is because if you draw the graph it's symmetric around 0.
The other rule that you can use is that P(Z>z) = 1P(Z<z), which you can kind of mean as (if it's not less than z then it's more than z or vice versa). Just like 'if the probability that it's going to rain is 0.2, then the probability that it won't rain is 0.8 (10.2)'.
So you've tried to use both of those things but you've got to keep track of how you do them. 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 8
 30102015 18:19
(Original post by SeanFM)
Good start
So you have P(2 < Z < 1.33), which can be written as P(Z<1.33)  P(Z<2). Why? You may wish to use a number line to help you.
You can split the range Z < 1.33 into two different regions: 2 < Z < 1.33 (which is what we want) and the rest is Z<2, and combining those two regions together again gives Z <  1.33. So if we express P(2 < Z < 1.33) in terms of P(Z <  1.33) then we would be counting the area 'Z < 2' but we're not interested in that, only the region 2 < Z < 1.33, so that's why it can be written as P(Z<1.33)  P(Z<2). That's just for your information/understanding.
Now the problem is that the tables have values P(Z<z) where z is a positive number, and you have to use properties of the normal distribution when you're dealing with things like P(Z>z) or P(Z<z) (where z is a positive number) or even P(Z>z).
So we've got P(Z<1.33)  P(Z<2) to work out. We need everything to be of the form P(Z<z), where z is a positive number.
So you use two things (one of which you've done already). P(Z<z) = P(Z>z) by symmetry. This is because if you draw the graph it's symmetric around 0.
The other rule that you can use is that P(Z>z) = 1P(Z<z), which you can kind of mean as (if it's not less than z then it's more than z or vice versa). Just like 'if the probability that it's going to rain is 0.2, then the probability that it won't rain is 0.8 (10.2)'.
So you've tried to use both of those things but you've got to keep track of how you do them. 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 9
 30102015 18:27
(Original post by SeanFM)
Good start
So you have P(2 < Z < 1.33), which can be written as P(Z<1.33)  P(Z<2). Why? You may wish to use a number line to help you.
You can split the range Z < 1.33 into two different regions: 2 < Z < 1.33 (which is what we want) and the rest is Z<2, and combining those two regions together again gives Z <  1.33. So if we express P(2 < Z < 1.33) in terms of P(Z <  1.33) then we would be counting the area 'Z < 2' but we're not interested in that, only the region 2 < Z < 1.33, so that's why it can be written as P(Z<1.33)  P(Z<2). That's just for your information/understanding.
Now the problem is that the tables have values P(Z<z) where z is a positive number, and you have to use properties of the normal distribution when you're dealing with things like P(Z>z) or P(Z<z) (where z is a positive number) or even P(Z>z).
So we've got P(Z<1.33)  P(Z<2) to work out. We need everything to be of the form P(Z<z), where z is a positive number.
So you use two things (one of which you've done already). P(Z<z) = P(Z>z) by symmetry. This is because if you draw the graph it's symmetric around 0.
The other rule that you can use is that P(Z>z) = 1P(Z<z), which you can kind of mean as (if it's not less than z then it's more than z or vice versa). Just like 'if the probability that it's going to rain is 0.2, then the probability that it won't rain is 0.8 (10.2)'.
So you've tried to use both of those things but you've got to keep track of how you do them.
The value of a if P(X<a) = 0.975
would I start by doing : P(Z < a  15 / 16 ) = 0.975?
of do I divide by 4 ?? 
Kevin De Bruyne
 Follow
 609 followers
 21 badges
 Send a private message to Kevin De Bruyne
 Very Important Poster
Offline21ReputationRep:Very Important Poster Follow
 10
 30102015 18:29
(Original post by Aty100)
Consider the distribution X~N(15,16) to calculate the following probability:
The value of a if P(X<a) = 0.975
would I start by doing : P(Z < a  15 / 16 ) = 0.975?
of do I divide by 4 ??
What next? 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 11
 30102015 18:32
(Original post by SeanFM)
Swap the 16 for 4 and you are correct remember that the second number is variance, and the square root of that is the standard deviation, so you could write it as N(15,(4^2)) just to remember that the standard deviation is 4.
What next?
a  15 / 4 = 0.975
so
a  15 = 3.9
add 15
a = 18.9 ??
sorry ive never done a question like this before 
Kevin De Bruyne
 Follow
 609 followers
 21 badges
 Send a private message to Kevin De Bruyne
 Very Important Poster
Offline21ReputationRep:Very Important Poster Follow
 12
 30102015 18:35
(Original post by Aty100)
the do u just do
a  15 / 4 = 0.975
so
a  15 = 3.9
add 15
a = 18.9 ??
sorry ive never done a question like this before
This time (in the opposite direction) we start with the probability, so what would we need to find out? 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 13
 30102015 18:49
(Original post by SeanFM)
Remember when we're working in the opposite direction, we start with an x value, turn it into a z value (so go from P(X<x) to P(Z<z), and find the probability that matches up with that z value.
This time (in the opposite direction) we start with the probability, so what would we need to find out?
Z = ??
I have no idea what I'm doing 
Kevin De Bruyne
 Follow
 609 followers
 21 badges
 Send a private message to Kevin De Bruyne
 Very Important Poster
Offline21ReputationRep:Very Important Poster Follow
 14
 30102015 18:54
Usually when we know what a is and we want to find out what the probability is (in this case 0.975 would be our final answer), so we take a, turn it into a z value using z = (amean)/standard deviation, then since we have P(Z<z) we go to the tables and find our answer, which would be 0.975 in this example.
But like I said last time we are working backwards... so we have the probability and we need to find out what a is. It is 'reversing' the steps, so what do you think we should do? 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 15
 30102015 18:58
(Original post by SeanFM)
So we've got that P(Z < (a  15) / 4 ) = 0.975
Usually when we know what a is and we want to find out what the probability is (in this case 0.975 would be our final answer), so we take a, turn it into a z value using z = (amean)/standard deviation, then since we have P(Z<z) we go to the tables and find our answer, which would be 0.975 in this example.
But like I said last time we are working backwards... so we have the probability and we need to find out what a is. It is 'reversing' the steps, so what do you think we should do? 
Kevin De Bruyne
 Follow
 609 followers
 21 badges
 Send a private message to Kevin De Bruyne
 Very Important Poster
Offline21ReputationRep:Very Important Poster Follow
 16
 30102015 19:05
(Original post by Aty100)
would we find 0.975 in the table to find z which is 1.96 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 17
 30102015 19:35
(Original post by SeanFM)
Correct so what does that z equal to?
So how to I work out a? 
Kevin De Bruyne
 Follow
 609 followers
 21 badges
 Send a private message to Kevin De Bruyne
 Very Important Poster
Offline21ReputationRep:Very Important Poster Follow
 18
 30102015 20:44
Think about how you calculate a given z and work backwards. 
Aty100
 Follow
 8 followers
 14 badges
 Send a private message to Aty100
 Thread Starter
Offline14ReputationRep: Follow
 19
 30102015 20:53
(Original post by SeanFM)
How will z=1.96 be related to a?
Think about how you calculate a given z and work backwards.
P(1.96 < a) = 0.975 ?! 
Kevin De Bruyne
 Follow
 609 followers
 21 badges
 Send a private message to Kevin De Bruyne
 Very Important Poster
Offline21ReputationRep:Very Important Poster Follow
 20
 30102015 20:57
Remember that with an a value when we want to find P(X<a) we do (amean)/standard deviation to turn a into z, and find P(Z<z).
If a was 18 for example and we wanted to find P(X<18) then we would do P(Z<(1815)/4) = P(Z<3/4).
What I was trying to hint at was that each value for a has a value for z that we calcuate, and in that example (a15)/4 = 3/4, and in your question it's ..... = 1.96.
 1
 2

Mathematics, Optimisation and Statistics
Imperial College

Mathematics and Professional Education
University of Stirling

Northumbria University

Mathematics and Environmental Hazards
University of Derby

Mathematics Operational Research and Statistics with a Year Abroad
Cardiff University

University of Oxford

Mathematics and Statistics (Equal)
University of York

Mathematics and Statistics (including placement year)
University of Essex

Mathematics and Statistics with a Year Abroad
Queen Mary University of London

Secondary Mathematics Education with QTS
Edge Hill University
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
 SherlockHolmes
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 rayquaza17
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Labrador99
 EmilySarah00