Question

The average weight an NFL player is 245.7 pounds with standard deviation of 34.5 pounds but probability distribution of the population is unknown. If a random sample of 32 NFL players is selected, what is the probability that:
(i) the average weight of the sample will be less than 234 pounds?

(ii) the average weight of the sample is between 248-254?

(iii) the average weight of the sample is between 242-251?

Answer 1

Answer: For part I, the probability is 36.69% that the average weight will be less than 234.

To find the percents for each one, you need to find the z-score and use a normal distribution table. A z-score is the number of standard deviations above or below the mean.

For 234, the z-score would be:  (234 - 245.7) / 34.5 = -0.34 which corresponds to a percent of 36.69%.

For that last 2 parts, you follow the same plan. However, you will find two different percents. You will have to subtract them to find the percent between them.

For Part II, you should get 6.85%

For Part III, you should get 10.37%.