Stat cie question , please help me s1

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Universecolors
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#1
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#1
The weights of a letter posted by a certain business are normally distributed with mean 20g. It is found that the eights of 94% of the letters are within 12g of the mean .

1)find the standard deviation of the weights of the letters

My working:
8<X<32= 0.94
We have to convert the 0.94 to a z value right ? And I don’t know why it is not like that in the mark scheme

Here is the mark schem answer
Z=1.88
1.88=32-20/standard deviation

Answer=0.864
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Universecolors
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#2
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#2
The weights of a letter posted by a certain business are normally distributed with mean 20g. It is found that the eights of 94% of the letters are within 12g of the mean .

1)find the standard deviation of the weights of the letters

My working:
8<X<32= 0.94
We have to convert the 0.94 to a z value right ? And I don’t know why it is not like that in the mark scheme

Here is the mark schem answer
Z=1.88
1.88=32-20/standard deviation

Answer=0.864
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username1732133
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#3
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#3
(Original post by Universecolors)
The weights of a letter posted by a certain business are normally distributed with mean 20g. It is found that the eights of 94% of the letters are within 12g of the mean .

1)find the standard deviation of the weights of the letters

My working:
8<X<32= 0.94
We have to convert the 0.94 to a z value right ? And I don’t know why it is not like that in the mark scheme

Here is the mark schem answer
Z=1.88
1.88=32-20/standard deviation

Answer=0.864

What is the probability that the weight is greater than 32 or less than 8?

What is the probability that the weight is greater than 32?

What is the probability that the weight is less than 32?

Look for this probability in the table.

You will find that this probability = P(Z<1.88).

You might find drawing a sketch and writing in the probabilities for various areas helpful.
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Universecolors
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#4
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#4
(Original post by BuryMathsTutor)
What is the probability that the weight is greater than 32 or less than 8?

What is the probability that the weight is greater than 32?

What is the probability that the weight is less than 32?

Look for this probability in the table.

You will find that this probability = P(Z<1.88).

You might find drawing a sketch and writing in the probabilities for various areas helpful.
I don’t get it , when I put it in the calculator , u can’t do it , because u don’t have sd
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username1732133
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#5
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(Original post by Universecolors)
I don’t get it , when I put it in the calculator , u can’t do it , because u don’t have sd
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Universecolors
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#6
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#6
(Original post by BuryMathsTutor)
..
Thank u , I will try to go through the question again .
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RDKGames
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#7
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#7
(Original post by Universecolors)
The weights of a letter posted by a certain business are normally distributed with mean 20g. It is found that the eights of 94% of the letters are within 12g of the mean .

1)find the standard deviation of the weights of the letters

My working:
8<X<32= 0.94
We have to convert the 0.94 to a z value right ? And I don’t know why it is not like that in the mark scheme

Here is the mark schem answer
Z=1.88
1.88=32-20/standard deviation

Answer=0.864
You have P(8&lt;X&lt;32) = 0.94. Now the z values on your table are responsible for when you have P(Z&lt;z) therefore we need P(X&lt;x) first instead. So, you would prefer to get something of this form from what you already have. Indeed, using symmetry you can deduce that P(\mu&lt;X&lt;32) = 0.47 which is P(X&lt;32) - P(X&lt;\mu) = 0.47 and so P(X&lt;32) = 0.97.

Secondly, this is where the mark scheme comes in. Standardising X means that X \mapsto \dfrac{X-\mu}{\sigma} = Z. Hence we get that P(Z &lt; \frac{32-20}{\sigma}) = 0.97.
Hence \frac{32-20}{\sigma} = \Phi^{-1}(0.97) and from your tables, z=\Phi^{-1}(0.97) = 1.88.
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Universecolors
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#8
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#8
(Original post by BuryMathsTutor)
..
I really don’t get what u meAn by the graph
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Universecolors
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#9
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#9
(Original post by RDKGames)
You have P(8&lt;X&lt;32) = 0.94. Now the z values on your table are responsible for when you have P(Z&lt;z) therefore we need P(X&lt;x) first instead. So, you would prefer to get something of this form from what you already have. Indeed, using symmetry you can deduce that P(\mu&lt;X&lt;32) = 0.47 which is P(X&lt;32) - P(X&lt;\mu) = 0.47 and so P(X&lt;32) = 0.97.

Secondly, this is where the mark scheme comes in. Standardising X means that X \mapsto \dfrac{X-\mu}{\sigma} = Z. Hence we get that P(Z &lt; \frac{32-20}{\sigma}) = 0.97.
Hence \frac{32-20}{\sigma} = \Phi^{-1}(0.97) and from your tables, z=\Phi^{-1}(0.97) = 1.88.
Is more than and less than 12 taken into account in this way , don’t we have to multiply by 2
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RDKGames
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#10
(Original post by Universecolors)
Is more than and less than 12 taken into account in this way , don’t we have to multiply by 2
We don't get >12 or <12 from anywhere. The 12 is just how far away the ends of the interval must be from the mean. Not sure what you mean by multiplying by 2. Where? Perhaps you're thinking of P(\mu&lt; X &lt; 32) = 2 P(8 &lt; X&lt; 32) which is what we use in the working.

Also I do not recommend you using the TSR app on your phone because it doesn't display maths properly and you're probably seeing code in my replies which isn't supposed to appear like that.
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Universecolors
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#11
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(Original post by RDKGames)
We don't get >12 or <12 from anywhere. The 12 is just how far away the ends of the interval must be from the mean. Not sure what you mean by multiplying by 2. Where? Perhaps you're thinking of P(\mu&lt; X &lt; 32) = 2 P(8 &lt; X&lt; 32) which is what we use in the working.
What do you mean by /mu ?
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RDKGames
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(Original post by Universecolors)
What do you mean by /mu ?
Also I do not recommend you using the TSR app on your phone because it doesn't display maths properly and you're probably seeing code in my replies which isn't supposed to appear like that.

mu is the population mean. In this case it's 20.
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RDKGames
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#13
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#13
Please don't make multiple threads for a single question - there is no effect in doing so other than cluttering the forum!
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Golden313
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#14
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#14
What symmetry are you talking about, I'm stuck there pliz help
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davros
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#15
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#15
(Original post by Golden313)
What symmetry are you talking about, I'm stuck there pliz help
This thread is 4 years old so it would be better to start your own new thread to discuss the question, but very briefly the Normal distribution is symmetrical about the mean so if you draw a graph to represent the area (probability) in question then you should be able to see what's going on
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Golden313
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#16
OK thank you
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