Many firms use on-the-job training to teach their employees computer programming, suppose you work in the personnel department of a firm that just finished training a group of its employees to program, and you have been requested to review the performance of one of the trainees on the final test that was given to all trainees. the mean and standard deviation of the test scores are 80 and 5, respectively. assuming nothing is known about the distribution, what percentage of test-takers scored better than the trainee who scored 90? answer 1 point lkeypadtables at least 75% at least 25% at most 25% approximately all none of the above

  At most 25%

Step-by-step explanation:

The fraction of the distribution that can exceed n standard deviations from the mean is 1/n². For a standard deviation of 5 and a mean of 80, a score of 90 is (90-80)/5 = 2 standard deviations from the mean. Scores better than that can be at most (1/2)² = 1/4 = 25% of the total number of scores.

step-by-step explanation:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

step-by-step explanation:

given that megan owns a clothing boutique.

the profits of 2010, 11, are recorded and shown as follows:

years since     2010 0     1     2     3       4       5

profits made in 1000's   39       84     111     120   111     84

we find that profit gradually increases form 2010 to 2013 upto 120,   then starts decreasing.

the profits for 2012 and 2014 are the same while that of 2011 and 2015 are also the same.

this shows that there is symmetry about the line profit = 120000

hence answer is:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

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step-by-step explanation: