Moivre-Laplace theorem (homework)

[Poetria]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

I have no idea what I am doing wrong. Here is my problem:

S(n) - the number of successes in 10 Bernoulli trials, each with probability 0.7

I am supposed to use normal approximation to compute :

P{|S(n)/10-0.7|>0.2}

We are allowed to use a calculator.

I got:
|S(n)-7|>2|
5<S(n)>9

I have computed S*:

S(n)-7/1.44914
7=mean
1.44914- standard deviation

5*1.44914-7<Z<9*1.44914-7

Area (probability): 0.2451

But there is no such answer in the options.

 

[SammyS]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

I have no idea what I am doing wrong. Here is my problem:

S(n) - the number of successes in 10 Bernoulli trials, each with probability 0.7

I am supposed to use normal approximation to compute :

P{|S(n)/10-0.7|>0.2}

We are allowed to use a calculator.

I got:
|S(n)-7|>2|
5<S(n)>9

I have computed S*:

S(n)-7/1.44914
7=mean
1.44914- standard deviation

5*1.44914-7<Z<9*1.44914-7

Area (probability): 0.2451

But there is no such answer in the options.

What do you mean by 5<S(n)>9 ?

That's an invalid compound inequality.

 

[Poetria]

I see. :( My line of thought was as follows:

S(n)-7>2 or S(n)-7<-2 (this is an absolute value equation: |S(n)/10-0.7|>0.2)

How to correct it?

 

[SammyS]

I see. :( My line of thought was as follows:

S(n)-7>2 or S(n)-7<-2 (this is an absolute value equation: |S(n)/10-0.7|>0.2)

How to correct it?

What you have here is correct.

S(n)-7>2 or S(n)-7<-2

That says S(n) > 9 or S(n) < 5 . (You had S(n) > 5 )

You can't make these into a compound inequality.

 

[Ray Vickson]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

I have no idea what I am doing wrong. Here is my problem:

S(n) - the number of successes in 10 Bernoulli trials, each with probability 0.7

I am supposed to use normal approximation to compute :

P{|S(n)/10-0.7|>0.2}

We are allowed to use a calculator.

I got:
|S(n)-7|>2|
5<S(n)>9

I have computed S*:

S(n)-7/1.44914
********************
Wrong: use parentheses, like this: (S(n) - 7)/1.44914

********************
7=mean
1.44914- standard deviation

5*1.44914-7<Z<9*1.44914-7

***********************
Wrong: you need 5 > 7 + 1.44914*Z or 7 + 1.44914*Z > 9

The complement of the event "{S(n) < 5} or {S(n) > 9}" is {5 <= S(n) <= 9}, which translates to
5 <= 7 + 1.44914*Z <= 9, or -1.38 <= Z <= 1.38 in the normal approximation.

However, since n = 10 is small, the normal approximation might not be very good, but it can be improved a lot by using the so-called "1/2-correction". The random variable S(n) is discrete (taking values 0,1,2,...,10 only), so {5 < S(n) < 9} = {S(n) = 6,7,8}. In the normal approximation, replace this by {5.5 <= S_normal <= 8.5}, so
5.5 <= 7 + 1.44914*Z <= 8.5. The resulting increase in accuracy is striking in this example.

*********************
Area (probability): 0.2451

But there is no such answer in the options.

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

I have no idea what I am doing wrong. Here is my problem:

S(n) - the number of successes in 10 Bernoulli trials, each with probability 0.7

I am supposed to use normal approximation to compute :

P{|S(n)/10-0.7|>0.2}

We are allowed to use a calculator.

I got:
|S(n)-7|>2|
5<S(n)>9

I have computed S*:

S(n)-7/1.44914
7=mean
1.44914- standard deviation

5*1.44914-7<Z<9*1.44914-7

Area (probability): 0.2451

But there is no such answer in the options.

 

[Poetria]

Thank you so much. I have tried to work it out. I think I got it. My problem is that I still can't get an exact answer. The closest (with the correction you have suggested):
if x = 5.5 I get 0.15031188 (I can choose 0.1675)
http://www.danielsoper.com/statcalc3/calc.aspx?id=53

The result: 0.08377348 is very imprecise.
Actually some calculators demand x, some Z. It is a mess.

 

[Poetria]

Sorry, I get 0.71860226. Even worse.

 

[Poetria]

I got it at last. :) Many thanks. :) You have helped me a lot. :) :) :)