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S1 normal distribution!
I know all the basics,e.g. how to find X when given the mean and standard deviation.But can someone explain
(a) how exactly do you find the mean and standard deviation from the normal distribution,when they are NOT given?!Please could someone go through it step by step?
(b) Also,i understand that sometimes you have to look up the z value in the percentage points table (for some reason?!...).But sometimes in mark schemes they use the table to find a value,but use the negative sign infront of it?!anyone know why and when you change the sign?!
finally,help would be appreciated if someone can run me through this question...
From experience a high jumper knows that he can clear a height of at least 1.78 m once in 5 attempts. He also knows that he can clear a height of at least 1.65 m on 7 out of 10 attempts.
Assuming that the heights the high jumper can reach follow a Normal distribution,
(a) draw a sketch to illustrate the above information, (3marks)
(b) find, to 3 decimal places, the mean and the standard deviation of the heights the high jumper can reach, (6marks)
(c) calculate the probability that he can jump at least 1.74 m. (3marks)
I can do (a) its just (b) and (c) that i cannot do,and don't understand how to do.
(a) how exactly do you find the mean and standard deviation from the normal distribution,when they are NOT given?!Please could someone go through it step by step?
(b) Also,i understand that sometimes you have to look up the z value in the percentage points table (for some reason?!...).But sometimes in mark schemes they use the table to find a value,but use the negative sign infront of it?!anyone know why and when you change the sign?!
finally,help would be appreciated if someone can run me through this question...
From experience a high jumper knows that he can clear a height of at least 1.78 m once in 5 attempts. He also knows that he can clear a height of at least 1.65 m on 7 out of 10 attempts.
Assuming that the heights the high jumper can reach follow a Normal distribution,
(a) draw a sketch to illustrate the above information, (3marks)
(b) find, to 3 decimal places, the mean and the standard deviation of the heights the high jumper can reach, (6marks)
(c) calculate the probability that he can jump at least 1.74 m. (3marks)
I can do (a) its just (b) and (c) that i cannot do,and don't understand how to do.
Reply 1
From experience a high jumper knows that he can clear a height of at least 1.78 m once in 5 attempts. He also knows that he can clear a height of at least 1.65 m on 7 out of 10 attempts.
Assuming that the heights the high jumper can reach follow a Normal distribution,
(b) find, to 3 decimal places, the mean and the standard deviation of the heights the high jumper can reach, (6marks)
Height he can jump are distributed normally with mean μ and variance σ² as described here:
J~N(μ,σ²)
we also know that
P(J>1.78) = 0.20 and
P(J>1.65) = 0.70
now we have to standardise to get these probabilities in terms of Z so we can look up the probablilties in the table given in the formula book:
so we take away the mean and divide by the standard deviation:
P(Z>(1.78-μ
/σ = 0.20
P(Z>(1.65-μ/σ = 0.70
therefore:
Ф((1.78-μ/σ = 1-0.20
= 0.80 so from the table
(1.78-μ/σ = 0.7881
Ф((1.65-μ/σ = 1-0.70
= 0.30 so from the table
(1.65-μ/σ = 0.6179
so now we have two equations with two lots of μ and σ in them, which we can solve for μ and σ:
1.78-μ = 0.7881σ
1.65-μ = 0.6179σ
i hope you can follow it and i hope i did it right!
if you're stuck, theres an example on page 174 of the blue heinemann S1 book (its Example 3, some books vary in their page numbers) i just followed that example which is similar to the question.
Assuming that the heights the high jumper can reach follow a Normal distribution,
(b) find, to 3 decimal places, the mean and the standard deviation of the heights the high jumper can reach, (6marks)
Height he can jump are distributed normally with mean μ and variance σ² as described here:
J~N(μ,σ²)
we also know that
P(J>1.78) = 0.20 and
P(J>1.65) = 0.70
now we have to standardise to get these probabilities in terms of Z so we can look up the probablilties in the table given in the formula book:
so we take away the mean and divide by the standard deviation:
P(Z>(1.78-μ
/σ = 0.20P(Z>(1.65-μ
/σ = 0.70therefore:
Ф((1.78-μ
/σ = 1-0.20= 0.80 so from the table
(1.78-μ
/σ = 0.7881Ф((1.65-μ
/σ = 1-0.70= 0.30 so from the table
(1.65-μ
/σ = 0.6179so now we have two equations with two lots of μ and σ in them, which we can solve for μ and σ:
1.78-μ = 0.7881σ
1.65-μ = 0.6179σ
i hope you can follow it and i hope i did it right!
if you're stuck, theres an example on page 174 of the blue heinemann S1 book (its Example 3, some books vary in their page numbers) i just followed that example which is similar to the question.
Thanks a lot for your help,but that is not the right answer.the two equations from the mark scheme were: 1.78-μ = 0.8416σ and 1.65-μ = -0.5244σ
I actually have NO IDEA how they get these answers,but they may be from percentage points table,however you use that?!please could someone find out how to do this?thanks
I actually have NO IDEA how they get these answers,but they may be from percentage points table,however you use that?!please could someone find out how to do this?thanks
Reply 3
oh these ones come from the smaller table.
i thought about using that table but it didnt have 0.7 on it so i just did 1-the probabilities and thought that would work
if you look they used the values 0.2 and 0.3 (which is 1-0.7) to get those values.
i thought about using that table but it didnt have 0.7 on it so i just did 1-the probabilities and thought that would work
if you look they used the values 0.2 and 0.3 (which is 1-0.7) to get those values.
Reply 4
Can i ask how the graph should look?
Reply 5
If both are more than why does graph look like this
http://img399.imageshack.us/my.php?image=s1ja1.jpg
http://img399.imageshack.us/my.php?image=s1ja1.jpg
Reply 7
insparato
If P( x > 1.65) = 0.7
that means area below 1.65 = 0.3
Therefore P( x < 1.65) = 0.3 which is shown on the graph.
that means area below 1.65 = 0.3
Therefore P( x < 1.65) = 0.3 which is shown on the graph.
Makes sense now thanks
Reply 8
bekky92
The Percentage Points Table
If you get a question that gives you a percentage chance of something happening, but no values for the mean and S.D, you can use this table to find the value of z for which there is the variable Z exceeds with these percentages.
Say, there is a 20% chance that somebody is over 1.75 metres
P(X>1.75) = 0.2
First, you standardise it.
P(Z>(1.75-μ/σ = 0.2
Then you find 0.2 in the table (under the column titled "p")
The corresponding z value is on the column to the right, titled "z".
For p = 0.2, z = 0.8416
This means that (1.75-μ/σ=0.8416
So 1.75-μ=0.8416σ
the thing to watch out for with these tables, is that Z must be > that something else [so P(Z>z)], otherwise it won't work
If you get a question that gives you a percentage chance of something happening, but no values for the mean and S.D, you can use this table to find the value of z for which there is the variable Z exceeds with these percentages.
Say, there is a 20% chance that somebody is over 1.75 metres
P(X>1.75) = 0.2
First, you standardise it.
P(Z>(1.75-μ
/σ = 0.2Then you find 0.2 in the table (under the column titled "p")
The corresponding z value is on the column to the right, titled "z".
For p = 0.2, z = 0.8416
This means that (1.75-μ
/σ=0.8416So 1.75-μ=0.8416σ
the thing to watch out for with these tables, is that Z must be > that something else [so P(Z>z)], otherwise it won't work
i've been wondering about this for ages, thanks so much!
is it possible to work out these questions with the big table though? because before i knew what the percentage points table was for, i used the big one (backwards) to do the question itsme!! was asking about and ended up with the right answer
, was that just a fluke? thanksReply 9
xX_JIMI_Xx
Ф((1.65-μ/σ = 1-0.70
= 0.30 so from the table
(1.65-μ/σ = 0.6179
/σ = 1-0.70= 0.30 so from the table
(1.65-μ
/σ = 0.6179You are completely and utterly incorrect.
You have P(Z < (1.65-u)/o) = 0.3
so Ф((1.65-u)/o) = 0.3
There is no 0.3 value for Ф(z) on the table. There is a 0.3 value for z, but you haven't got z = 0.3, you've got Ф(z) = 0.3.
To correctly answer this part of the question, you would do the following:
P(Z < (1.65-u/o)) = 0.3
Ф((1.65-u)/o) = 0.3
(1.65-u)/o = [Ф to the power of -1]0.3
Since you can't look up 0.3 in the tables you have to multiply your z value by -1 and do 1 - the probability.
So
(u-1.65)/o = [Ф to the power of -1]0.7
Then you can look up 0.7 on the normal table or 0.3 on the more accurate table and you have either
(u-1.65)/o = 0.52
or, more accurately:
(u-1.65)/o = 0.5244
In the future please do not post advice on a topic which you do not fully understand. The inquirer probably got a worse mark because of you.
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