a. 68% of the marks on a test are between 81 and 94. Assume the data is normally distributed, what are the mean and standard deviation?

Mean =

Standard Deviation=

b. This dealt with the weights of your football team.The average weight was 226.9 with a standard deviation of 46.54.

The coach has heard that a new ball player will be transferring in the Fall.He has heard three different weights for the new player; 150, 180, and 350.

Find the z-scores for these three weights and determine from them which is the least likely weight of the new player.Explain your answer.

Weight 150: The Z score is

Weight 180: The Z score is

Weight 350: The Z score is

Determine which is the least likely weight of the new player:

Explain your answer:

c. This dealt with the weights of your football team.The average weight was 226.9 with a standard deviation of 46.54.

The coach has heard that a new ball player will be transferring in the Fall.He has heard three different weights for the new player; 150, 180, and 350.

Find the z-scores for these three weights and determine from them which is the least likely weight of the new player.Explain your answer.

Weight 150: The Z score is

Weight 180: The Z score is

Weight 350: The Z score is

Determine which is the least likely weight of the new player:

Explain your answer:

d. A very good example of the value of z-scores is the comparison a college must make between ACT scores and SAT scores when is determining admission requirements.These two tests are comparable and, with many colleges, carry equal weights.However, the average combined SAT score is 1500 with a standard deviation of 250, while, for the ACT, these numbers are 20.8 and 4.8.

Suppose, two potential candidates for admission are equal except that one has scored 27 on the ACT and the other, 1800 on the SAT.Which one has the better test score?

di.Calculate the z-scores of the two students and determine who had the better scores.

Z score for Student 1:

Z score for Student 2:

Determine who had the better scores:

dii.Using one of the three methods of this section, calculate the area under the normal curve for each of these students and thereby their score's percentile (for example, in the top 20%).

For Student 1:

For Student 2: