Originally Posted byaptiva

Mean score:

.

RonL

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- May 25th 2006, 07:36 PM #1

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## 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]

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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 26th 2006, 12:13 AM #2

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- May 26th 2006, 12:34 AM #3

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Originally Posted by**aptiva**

distribution are .

The mean of a sample of 200 drawn from a population with mean

is . The standard deviation of the mean of a sample of 200 drawn

from a population with standard deviation of is .

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

and:

So the required interval is

b) Construct a 99% confidence interval to estimate the mean weekly expenditure on food in the population. [83.96 to 86.92]

z-scores for a 99% region.

RonL