A random sample of 18 graduates of a certain secretarial school typed an average of 80.6 words per minute with a standard deviation of 7.2 words per minute. Assuming a normal distribution for the number of words typed per minute, compute the 95% prediction interval for the next observed number of words per minute typed by a graduate of the secretarial school.

( 77.27 , 83.93)

Therefore at 95% confidence/prediction interval is

= ( 77.27 , 83.93)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 80.6 words per minute

Standard deviation r = 7.2

Number of samples n = 18

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

80.6+/-1.96(7.2/√18)

80.6+/-1.96(1.697056274847)

80.6 +/- 3.33

= ( 77.27 , 83.93)

Therefore at 95% confidence/prediction interval is

= ( 77.27 , 83.93)

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

i think the amount would be $140.62