Statistic Questions?

Question

1.

Coke cans are filled with 12 ounces of soda with a standard deviation of 0.11 oz. Find the probability that in a sample of 49 cans, the mean will be greater than 12.10 ounces (profit waste).

2.
What conditions must be true in order to use the normal distribution to approximate the binomial distribution?

3.
Based on observed males using public restroom, 85% of adult males wash their hands in public. In a survey of 523 males, 518 reported they wash their hands in public restrooms. Assuming the 85% rate is correct, find the probability that among 523 males, 518 or more wash their hands in public restrooms. What do you conclude with this result?

4.
For the data values 45, 63, 28 and 39, find the mean of the sample means and the standard deviation of the sample means for all possible samples of size 3.

5.
Of U.S. households, 67% have Internet access. What is the probability that in a sample of 400 households, between 300 and 350 will have Internet access?

Answer 1

12 ounce cans, standard deviation of .11 ounce

sigma = .11, mu = 12, N=49

what is the probability in a sample of 49 cans, the sample mean will be greater than 12.1 ounces?

sample deviation = s = sigma/sqrN = .11/sqr49 = .11/7 = about .0157

find the z score for 12.1

(12.1-12)/.0157 = about .1/.0157 = 6.363636...standard deviations from the mean

use a z calculator or z tables to find the corresponding area under the curve

answer is virtually zero = .0000.... with a lot of zeros before getting a non zero digit

round off to more decimals than anyone ever rounds off and you still get 0% probability of more than 12.1 ounces = the sample mean

to use the normal distribution as the approximation for the sample, the sample size should be "large," such as greater than 30. to use the normal distribution to approximate the binomial distribution, np and nq should be greater than 5, even better if they are greater than 10. then you can approximate the mean with np and the standard deviation as square root of npq or np(1-q)

67% have internet access. in a sample of 400 households what is the probability between 300 and 350 have internet access?

.67 x 400 = 268 households.

find the z scores for 300 and 350

(300-268)/s= 32/s=z for 300, where s= the standard deviation

(350-268)/s = 82/s=z for 350

find the areas under the normal curve for the two z scores and subtract them. that gives you the probability of between 300 and 350