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Heckler
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#1
Could someone please help in solving this problem for me, could you also include working or explain
how you got the answer. thanks Question is as follows:

"45% of people in a population use Shine toothpaste. A telephone survey is done and 400 people asked
which brand of toothpaste they use. The probability that more than one half use Shine is what?

1) 4.8%
2) 4.0%
3) 2.2%
4) 4.6%
5) 50% "

any help would be great please

thanks
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Adam Atkinson
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On 12-May-02 05:53:13, Heckler said:
[q1]>Could someone please help in solving this problem for me, could you also include working or explain[/q1]
[q1]>how you got the answer. thanks[/q1]

[q1]>"45% of people in a population use Shine toothpaste. A telephone survey is done and 400[/q1]
[q1]>people asked which brand of toothpaste they use. The probability that more than one half use[/q1]
[q1]>Shine is what?[/q1]

[q1]>1) 4.8%[/q1]
[q1]>2) 4.0%[/q1]
[q1]>3) 2.2%[/q1]
[q1]>4) 4.6%[/q1]
[q1]>5) 50%[/q1]

Looks to me like a question which expects you to use the normal approximation to the binomial
distribution.

--
Adam Atkinson ([email protected]) "Let's catch that sick bird" he said, illegally.
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Gareth Jones
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[email protected] (Heckler) wrote:

[q1]>Could someone please help in solving this problem for me, could you also include working or explain[/q1]
[q1]>how you got the answer. thanks Question is as follows:[/q1]
[q1]>[/q1]
[q1]>"45% of people in a population use Shine toothpaste. A telephone survey is done and 400[/q1]
[q1]>people asked which brand of toothpaste they use. The probability that more than one half use[/q1]
[q1]>Shine is what?[/q1]

The probability that r people will use Shine is:
p(r) = nCr * p^r * (1-p)^(n-r)

where p = 0.45, n = 400.

This is the binominal distribution.

You need to add p(200) + p(201) + .... + p(400), which is quite hard.

Fortunately, you can approximate a binominal distribution with a normal distribution with mean np
and variance np(1-p) ie. N(180, 99).

So you want to know what is the probability that r > 200 in the distribution N(180, 99). That is the
same as the probability that

q > (200-180)/sqrt(99) in the distribution N(0,1).

q.a. that z > 2.01.

You can lookup this probability in a table of the standard normal distribution, and it comes to
0.022, i.e. 2.2%.

Gareth

[q1]>1) 4.8%[/q1]
[q1]>2) 4.0%[/q1]
[q1]>3) 2.2%[/q1]
[q1]>4) 4.6%[/q1]
[q1]>5) 50% "[/q1]
[q1]>[/q1]
[q1]>any help would be great please[/q1]
[q1]>[/q1]
[q1]>thanks[/q1]
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Adam Atkinson
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On 12-May-02 22:21:04, Gareth Jones said:

[q1]>So you want to know what is the probability that r > 200 in the distribution N(180, 99). That is[/q1]
[q1]>the same as the probability that[/q1]

[q1]>z > (200-180)/sqrt(99) in the distribution N(0,1).[/q1]

It may not make any difference here, but it's a good idea to remember to correct for continuity when
doing this sort of thing, isn't it? In this case I'd have used 200.5 instead of 200.

--
Adam Atkinson ([email protected]) If I were a fuzzy wuzzy bear 3 would be a perfect square
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