A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is the 90% confidence interval for the mean score, muμ, of all students taking the test?

29.5+/-1.11

= ( 28.39, 30.61)

Therefore, the 90% confidence interval (a,b) =( 28.39, 30.61)

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 = 29.5

Standard deviation r = 5.2

Number of samples n = 59

Confidence interval = 90%

z-value (at 90% confidence) = 1.645

Substituting the values we have;

29.5+/-1.645(5.2/√59)

29.5+/-1.645(0.676982337100)

29.5+/-1.113635944529

29.5+/-1.11

= ( 28.39, 30.61)

Therefore, the 90% confidence interval (a,b) =( 28.39, 30.61)

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3