# need help with these Q's please

• May 25th 2006, 06:36 PM
aptiva
need help with these Q's please
Players select one chip froma bag with 8 zerosand 4 ones and one chip froma bag with 4 zeros and 8 ones. Their score is the larger of the two numbers on the chips.

Present the probability distribution of the scores that could be achieved by a player in one game and compute its mean. [mu = 0.78]

__________________________________________________ ______________

In 1979,a random sample of 200 families in a large city was found to spend an average of $85.44 per week on food, with a standard deviation of$8.12.

a) Construct a 90% confidence interval to estimate the mean weekly expenditure on food in the population. [84.50 to 86.38]

b) Construct a 99% confidence interval to estimate the mean weekly expenditure on food in the population. [83.96 to 86.92]
• May 25th 2006, 11:13 PM
CaptainBlack
Quote:

Originally Posted by aptiva
Players select one chip froma bag with 8 zerosand 4 ones and one chip froma bag with 4 zeros and 8 ones. Their score is the larger of the two numbers on the chips.

Present the probability distribution of the scores that could be achieved by a player in one game and compute its mean. [mu = 0.78]

$
p(0)=prob(\mbox{0 from bag 1})prob(\mbox{0 from bag 2})$
$=\frac{8}{12} \frac{4}{12}\approx 0.2222
$

$
p(1)=1-p(0)
$

Mean score:

$
\mu=0\times 0.2222+1 \times 0.7778
$
.

RonL
• May 25th 2006, 11:34 PM
CaptainBlack
Quote:

Originally Posted by aptiva

In 1979,a random sample of 200 families in a large city was found to spend an average of $85.44 per week on food, with a standard deviation of$8.12.

a) Construct a 90% confidence interval to estimate the mean weekly expenditure on food in the population. [84.50 to 86.38]

The critical z-scores for a symmetric interval containing 90% of a normal
distribution are $\pm 1.645$.

The mean of a sample of 200 drawn from a population with mean $\ 85.44$
is $\ 85.44$. The standard deviation of the mean of a sample of 200 drawn
from a population with standard deviation of $\ 8.12$ is $\ 8.12/\sqrt{200}=\ 0.57$.

Therefore the z-scores of the end points of the required interval are:

$
z_{lo}=(x_{lo}-85.44)/0.57=-1.645
$

and:

$
z_{hi}=(x_{hi}-85.44)/0.57=+1.645
$

So the required interval is $(x_{lo},x_{hi})=(84.50,86.38)$

Quote:

b) Construct a 99% confidence interval to estimate the mean weekly expenditure on food in the population. [83.96 to 86.92]
This part is exactly the same except you will be working with the critical
z-scores for a 99% region.

RonL