An srs of 430 high school seniors gained an average of ¯x=20.44 points in their second attempt at the sat mathematics exam. assume that the change in score has a normal distribution with standard deviation σ = 52.32. we want to estimate the mean change in score u in the population of all high school seniors, give a 95% confidence interval (a, b) for based on this sample.

( 15.49,25.39)

Therefore at 95% confidence interval (a,b) = ( 15.49,25.39)

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 gain _x = 20.44

Standard deviation r = 52.32

Number of samples n = 430

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

+20.44+/-1.96(52.32/√430)

+20.44+/-1.96(2.523)

+20.44+/-4.95

= ( 15.49,25.39)

Therefore at 95% confidence interval (a,b) = ( 15.49,25.39)

127

whether it's diameter, or radius, i believe if i remember correctly from school, every amount has to always add up to 180. so if you take 53, and starting adding, long story short, taking 127 and adding it to 53, gives you 180.