Problems applying the central limit theorem

[giddy]

So from what I understand, the central limit theorem allows us to calculate the probability of the mean of a number of independent observations of the same variable.

I probably have not understood something because I can't really solve any of the problems just based on the formula give.

Homework Statement

An unbiased dice(6 sides) is thrown once. From its distribution its mean is 3.5 and vriable is 35/12. The same dice is thrown 70 times. 1. Find the probability that the mean score is less than 3.3. 2. Find the probability that the total score exceeds 260.

Homework Equations

<br /> X \sim N(\mu,\frac{\sigma^2}{n})<br />

The Attempt at a Solution

<br /> = X \sim N(3.5,\frac{35/12}{70})
So I have to find P(X<3.3) standardize X to Z
<br /> P(Z &lt; \frac{3.5-3.3}{\sqrt{0.0416}}) <br /> = P(Z &lt; 0.9806)<br /> =\phi(0.9806) = 0.8365<br />
And the answer is supposed to be 0.155!

I'm not sure at all how to approach the second part of the problem!?

 

[HallsofIvy]

So from what I understand, the central limit theorem allows us to calculate the probability of the mean of a number of independent observations of the same variable.

Well, it says that if a large number, n, of events have any probability distribution but with mean \mu and standard deviation \sigma then the mean of those events will be approximately normally distributed with mean \mu and standard deviation \sigma/\sqrt{n} while their sum will be approximately normally distributed with mean n\mu and standard deviation \sqrt{n}\sigma.

I probably have not understood something because I can't really solve any of the problems just based on the formula give.

Homework Statement

An unbiased dice(6 sides) is thrown once. From its distribution its mean is 3.5 and vriable is 35/12.

Do you mean the variance is 35/12?

The same dice is thrown 70 times. 1. Find the probability that the mean score is less than 3.3. 2. Find the probability that the total score exceeds 260.

Homework Equations

<br /> X \sim N(\mu,\frac{\sigma^2}{n})<br />

The Attempt at a Solution

<br /> = X \sim N(3.5,\frac{35/12}{70})
So I have to find P(X<3.3) standardize X to Z
<br /> P(Z &lt; \frac{3.5-3.3}{\sqrt{0.0416}}) <br /> = P(Z &lt; 0.9806)<br /> =\phi(0.9806) = 0.8365<br />
And the answer is supposed to be 0.155!

I'm not sure at all how to approach the second part of the problem!?

Where did you get \sqrt{0.0416}? If 35/12 is the variance of a single roll, then the variance for normal approximation is (35/12)/70= 1/24 and its standard deviation is \sqrt{1/24}= 0.2041,

 

[giddy]

hi.. yes I did mean variance... 1/24 = 0.41666...

Comes up close to the same thing... 3.5-3.3/0.2041 = 0.9799
P(Z < 0.9799) = 0.8368.. still not close to 0.155

 

[Dickfore]

In 1. The probablility you are asked is actually:

<br /> P(Z &lt; \frac{3.3-3.5}{\sqrt{\frac{35/12}{70}}}) = P(Z &lt; -0.980)<br />

I don't know what you mean by \phi(z)

 

[giddy]

<br /> \phi(z)<br />
That would be finding the probability after standardizing X to Z so Z ~ N(0,1) there's a table I have in the back of my book where i look up the probability then.

Thanks that's correct...but its a little off from the answer:
P(Z < -0.980)
= 1 - phi(0.980)
1 - 0.8363 = 0.1637
The answer in my book is 0.155 but I've learned from a lot of mistakes that the textbook is very accurate.

 

[Dickfore]

Thanks that's correct...but its a little off from the answer:
P(Z < -0.980)
= 1 - phi(0.980)
1 - 0.8363 = 0.1637
The answer in my book is 0.155 but I've learned from a lot of mistakes that the textbook is very accurate.

The mean of an unbiased dice is:

<br /> \mu = \mathrm{E}(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{7}{2}<br />

and the mean of the squares is:

<br /> \mathrm{E}(X^{2}) = \frac{1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6} = \frac{7 \cdot 13}{6}<br />

and the variance is:

<br /> \sigma^{2} = \mathrm{var}(X) = \mathrm{E}(X^{2}) - [\mathrm{E}(X)]^{2} = \frac{7 \cdot 13}{6} - \frac{7 \cdot 7}{4} = \frac{7 \cdot (2 \cdot 13 - 3 \cdot 7)}{12} = \frac{35}{12}<br />

so, there is no doubt about what you gave in the problem formulation. I checked in Mathematica that \Phi(0.980) = 0.836457. The complementary of that is 0.163543.

If the book is right, then the complementary of 0.155 is 0.845. The z-score corresponding to this is \Phi(z_{0} = 0.845 \Rightarrow z_{0} = 1.01522, which would mean that:

<br /> \frac{3.3 - \mu}{\sqrt{\frac{\sigma^{2}}{n}}} = -z_{0} \Rightarrow \frac{\sigma^{2}}{n} = \left(\frac{3.3 - \mu}{z_{0}}\right)^{2}<br />

<br /> n = \frac{z^{2}_{0} \, \sigma^{2}}{(3.3 - \mu)^{2}} = \frac{1.01522 \times 2.9167}{0.04} = 74<br />

 

[giddy]

hey, thanks so much. Seems the problem with the next 2 problems was post #4. But what about the second part of the problem... the next problem in my book is similar and I don't even know how to approach it.

A rectangular field is gridded into squares of side 1m. At one time of the year the number of snails in the field can be modeled by the Poission Distribution with a mean of 2.25 per m<sup>2.

Show that the probability of observing at least 200 snails in a random sample of 100 grid squares is approximately 95%.

So X ~ Po(2.25) E(X) = 2.25 and Var(X) = 2.25. So could you gimme a hint?</sup>

 

[Dickfore]

We are not a problem solving forum. You need to present your own work.

 

[giddy]

yea, I understand. I just wanted to know how to approach the problem. I've tried a few random things.

I just tried applying : <br /> X \sim N(\mu,\frac{\sigma^2}{n})<br /> for P( X > 200) = 200-2.25/\sqrt{2.25/100} = -1318.33 Which can't be standardized.
Also, trying T = probability of total number of snails
E(T) = 200* E(X) = 200 * 2.25
Var(T) = 200*Var(X) = 200 * (2.25/100)
T ~ N(450, 4.5)
P(T &gt; 200) = 200-450/\sqrt{4.5} = -117.8
Again a bogus answer.

I don't really know how to begin?

 

[vela]

You're getting screwed up answers because you're making numerous errors by mixing up quantities left and right. Try rereading the first paragraph of HallsofIvy's post above and then carefully apply the central limit theorem again. Remember you have two different distributions involved in the problem, one for a single lot, which is described by the Poisson distribution, and one for the total of 100 lots, which is described by the approximately Gaussian distribution.

 

[giddy]

Sorry, I can be very distracted(Was diagnosed with ADD symptoms but the meds didn't do much)

So the second part of what HallsofIvy's post says about the central limit theorem isn't in my book.The part where the sum of the probabilities is mean * n and the SD is sqrt{n * SD}.
So that seems to give my the answer 0.9522! 95%

I'm just a little bit on edge because for some reason every one of my answers to all the problems in this jinxed exercise is slightly off. The books answer to this problem is 0.9554 above I got 0.9522, the next problem the answer is 0.0082 whereas my answer is 0.0079. Another problem... I solved... for n with the central limit theorem and the answer was bogus so I solved it backward like dickfore did in #6 with the supposed answer for n and I get a a bogus probability. =P Normally these CIE endorsed books are very accurate to the last decimal.

That was the last of this chapter.. . I'm moving on to Chapter 5 - Estimation now... sorry for all the trouble! Thanks!