need help with these Q's please

**aptiva** · May 25th 2006, 06:36 PM

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]

**CaptainBlack** · May 25th 2006, 11:13 PM

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

**CaptainBlack** · May 25th 2006, 11:34 PM

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 $(x_{lo},x_{hi})=(84.50,86.38)$

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

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need help with these Q's please