# 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]

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

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

Mean score:

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

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 $\displaystyle \pm 1.645$.

The mean of a sample of 200 drawn from a population with mean $\displaystyle \$ 85.44$is$\displaystyle \$85.44$. The standard deviation of the mean of a sample of 200 drawn
from a population with standard deviation of $\displaystyle \$ 8.12$is$\displaystyle \$8.12/\sqrt{200}=\$ 0.57$. Therefore the z-scores of the end points of the required interval are:$\displaystyle
z_{lo}=(x_{lo}-85.44)/0.57=-1.645
$and:$\displaystyle
z_{hi}=(x_{hi}-85.44)/0.57=+1.645
$So the required interval is$\displaystyle (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