# Sampling

• July 14th 2006, 10:06 PM
kingkaisai2
Sampling
An unbiased dice is thrown once. The var(x)=35/12. The same dice is thrown 70 times.
a) Find the probability that the mean score is less than 3.3.
b)Find the probability that the total score exceeds 260.

• July 15th 2006, 01:07 AM
CaptainBlack
Quote:

Originally Posted by kingkaisai2
An unbiased dice is thrown once. The var(x)=35/12. The same dice is thrown 70 times.
a) Find the probability that the mean score is less than 3.3.
b)Find the probability that the total score exceeds 260.

I will assume that this is a question where we use the central limit theorem to
justify treating the distribution of the mean score as normal with mean equal
to the mean for a single throw of the die $\mu=3.5$ and variance equal to the variance
of the result of a single throw divided by the sample size $\sigma^2=(35/12)/70=1/24$.

a) Now a mean of less than 3.3 means 230 or less for the sum, so to correct
for the continuity of the normal approximation we use a z-score for a mean
of 230.5/70, which is the z-score for for a mean of 3.2928 so:

$
z=\frac{3.2928-3.5}{1/\sqrt{25}}\approx -1.015
$

Looking this up in a normal table or using a normal calculator we have for
a normal distributed RV we have:

$
p(z<-1.015) \approx 0.155
$

b) There are at least two equivalent ways to do this, one another is to
observe that the event: "total score exceeds 260.5" is the same as "the mean
exceeds 260.5/70". I will use the latter, then the z-score for a mean of
260.5/70 (again the extra 0.5s are continuity corrections) is:

$
z=\frac{260.5/70-3.5}{1/\sqrt{24}}\approx 1.0848
$

Looking this up as before we find:

$
p(z<1.0848) \approx 0.861
$

so:

$
p(z>1.0848) = 1-p(z<1.0848) \approx 0.139
$
.

RonL

Notes:
i. A mean of less than 3.3 means the sum of the 70 throws is less than
3.3x70=231. So as the normal distribution is continuous and the actual
distribution of the sum and mean is discrete we use in the normal model a
sum less than 230.5 to represent an actual sum of 230 or less.

ii. Similarly in part b), a total of more than 260 is modelled as more than 260.5
in the normal approximation.

iii. Given the nature of the way problems are set, there is a significant chance
that the expected answers to the two parts of this problem are the same.
So please check the working to see if there is some slight modification which
would make this so.