Thread: binomial

3. Suppose thirty-five percent of adults did not visit their physicians' offices last year. Let x be the number of adults in a random sample of 40 adults who did not visit their physicians' offices last year. For the probability distribution of x:
a] The mean is: _______ because: ______________________
b] The standard deviation is: ________ because: ____________________
NOTE: HINT: use the t statistic to answer problem 4 (you are working with a small sample, n=12)
4.Construct a 99% confidence interval for the mean of the population of all statistics students from which the sample in problem 1 was drawn.
5.According to the American Federation of Teachers, the average salary of 1998 college graduates with a degree in accounting was $33,702 (USA Today, June 18, 1999). Assume that the distribution of current salaries of recent graduates with a degree in accounting is skewed to the right with a mean of $33,702 and a standard deviation of $3,900. Find the probability that the mean current salary of a random sample of 200 recent college graduates with a degree in accounting is within $400 of the population mean.

thank you for any help and suggestion in advance.

Originally Posted by mathdummy

3. Suppose thirty-five percent of adults did not visit their physicians' offices last year. Let x be the number of adults in a random sample of 40 adults who did not visit their physicians' offices last year. For the probability distribution of x:
a] The mean is: _______ because: ______________________
b] The standard deviation is: ________ because: ____________________

This is a question on the application of the binomial distribution.

The number in the sample who did not visit the doctor last year is ~B(n;40,0.35).

The mean of X~B(n,N,p) is np, and the standard deviation is sqrt(Np(1-p)).

RonL

Originally Posted by mathdummy

NOTE: HINT: use the t statistic to answer problem 4 (you are working with a small sample, n=12)
4.Construct a 99% confidence interval for the mean of the population of all statistics students from which the sample in problem 1 was drawn.

Can't be done without the data from question 1.

RonL

Originally Posted by mathdummy

5.According to the American Federation of Teachers, the average salary of 1998 college graduates with a degree in accounting was $33,702 (USA Today, June 18, 1999). Assume that the distribution of current salaries of recent graduates with a degree in accounting is skewed to the right with a mean of $33,702 and a standard deviation of $3,900. Find the probability that the mean current salary of a random sample of 200 recent college graduates with a degree in accounting is within $400 of the population mean.

You are expected to use the result that a sample mean is approximatly normally distributed with mean equal to the population mean, and sd = population sd/sqrt(N), N the sample size, for large enough samples. (this is the Central Limit Theorem)

So assume the sample mean m~N(33900, (3900^2)/200), so now you need to find the probability p(-400<m-33900<400), or dividing through by 3900/sqrt(N) that:

p(-1.450475<z<1.450475), where z=[m-33900]/[3900/sqrt(200)], has a standard normal distribution.

RonL

1. The test grades for twelve students in a certain statistics class are:
87, 60, 72, 98, 99, 59, 67, 94, 92, 89, 93, 100
a] The mode is Not Available because there are no repetitive pattern within the number set.
b] The mean is 84.167 because this is the calculated result of the mean of this number set.
c] The std. dev. is 15.3435 (show the formula you used)
d] The median is 90.5 because this is the mid range of the number within this set.
e] The range is 41 because this is the highest value and least value gap within the set.
f] The variance is 235.424 because it is the square of the standard deviation.
g] The 35th percentile is _____ because___________________________________________ _
this is the data set for number 4 what information should i use to solve the problem?

wow thank you that was so helpful. really appreciate the pointers. well its a lot more than just pointers thanks