When dragons on planet Pern lay eggs, the eggs are either green or yellow. The biologists have observed over the years that 34% of the eggs are yellow, and the rest green. Next spring the lead scientist has permission to randomly select 51 of the dragon eggs to incubate. Consider all the possible samples of 51 dragon eggs. What is the mean number of yellow eggs in samples of 51 eggs? (that is the same as asking for the expected value of the number of yellow eggs) (Give answer correct to at least one decimal place.) mean =  What is the standard deviation in the number of yellow eggs in samples of size 51? (Give answer correct to at least one decimal place.) standard deviation =  What is the variance in the number of yellow eggs in samples of size 51?

Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2.   Step 2 of 3 :   Compute the value of the test statistic. Round your answer to two decimal places.

The University of Montana admission standards require students to have an ACT score of at least 22.We know that Montana ACT scores are normally distributed with a mean of 20.1 and standarddeviation of 4.  If we ask a random sample of 100 students who took the ACT, how many would be expected toqualify for admission to UM?

According to the U.S. Department of Transportation’s Air Travel Consumer Report, the nation’s 12 largest airlines recorded an on-time arrival percentage of 77.4% in 2001. Of interest is to estimate the mean delay time for the 22.6% of all flights that did not arrive on time during 2013. A simple random sample of 28 late arriving flights was selected, and the mean delay time of this sample of 28 flights was 14.2 minutes, with a sample standard deviation of s= 6.4 minutes. Use this information to calculate and interpret a 98% confidence interval for the mean delay time for all flights that did not arrive on time during 2013.