Three potential employees took an aptitude test. each person took a different version of the test. the scores are reported below. vincent got a score of 83.583.5; this version has a mean of 68.268.2 and a standard deviation of 99. kaitlyn got a score of 251.2251.2; this version has a mean of 238238 and a standard deviation of 2222. reyna got a score of 8.048.04; this version has a mean of 7.27.2 and a standard deviation of 0.70.7. if the company has only one position to fill and prefers to fill it with the applicant who performed best on the aptitude test, which of the applicants should be offered the job?

Vincent and Reyna.

Step-by-step explanation:

As z-score indicates that a data point is how many standard deviation above mean, so to find which of three applicant should be offered the job, let us find the z-score for each person using z-score formula.

, where,

= z-score,

= Random sample score,

= Mean,

= Standard deviation.

Therefore, Vincent's score on aptitude test is 1.7 standard deviation above mean.

Therefore, Kaitlyn's score on aptitude test is 0.6 standard deviation above mean.

Therefore, Reyna's score on aptitude test is 1.2 standard deviation above mean.

Since Vincent and Reyna has higher z-scores, therefore, they are further above mean than Kaitlyn. Therefore, Vincent and Reyna should be offered the job.

The price of 1 senior citizen ticket is $4

The price of 1 adult ticket is $7

The price of 1 child ticket is $5

Step-by-step explanation:

Assume that the costs of a senior ticket is $x , an adult ticket is $y and

a child ticket is $z

First day:

The school sold 4 senior citizen tickets, 2 adult tickets and 5 child

tickets for a total of  $55

∴ 4x + 2y + 5z = 55 ⇒ (1)

Second day:

The school sold 7 senior citizen tickets, 2 adult tickets  and 5 child

tickets for $67

∴ 7x + 2y + 5z = 67 ⇒ (2)

Third day:

The school sold 2 senior citizen tickets, 4 adult tickets  and 2 child

tickets for $46

∴ 2x + 4y + 2z = 46 ⇒ (3)

The number of adult tickets and the number of child tickets in the first

and second days are equal, then we can subtract equation (1) from

equation (2) to find x

Subtract equation (1) from equation (2)

∴ (7x - 4x) + (2y - 2y) + (5z - 5z) = 67 - 55

∴ 3x = 12

Divide both sides by 3

∴ x = 4

Substitute the value of x in equation (2)

∴ 7(4) + 2y + 5z = 67

∴ 28 + 2y + 5z = 67

Subtract 28 from both sides

∴ 2y + 5z = 39 ⇒ (4)

Substitute the value of x in equation (3)

∴ 2(4) + 4y + 2z = 46

∴ 8 + 4y + 2z = 46

Subtract 8 from both sides

∴ 4y + 2z = 38 ⇒ (5)

Now lets solve equations (4) and (5) to find y and z

Multiply equation (4) by -2 to eliminate y

∴ -4y - 10z = -78 ⇒ (6)

Add equations (5) and (6)

∴ -8z = -40

Divide both sides by -8

∴ z = 5

Substitute the value of z in equation (4) or (5)

∴ 2y + 5(5) = 39

∴ 2y + 25 = 39

Subtract 25 from both sides

∴ 2y = 14

Divide both sides by 2

∴ y = 7

The price of 1 senior citizen ticket is $4

The price of 1 adult ticket is $7

The price of 1 child ticket is $5