Arandom sample of 125 students is chosen from a population of 4,000 students. if the mean iq in the sample is 100 with a standard deviation of 8, what is the 98% confidence interval for the students' mean iq score?

98.33 - 101.67

Step-by-step explanation:

The given Sample mean n= 100  

Standard deviation

Mean IQ in the sample

Standard error of mean =

Standard error of mean =

Thus, standard error =

Standard error of mean (SE)= 0.7155  

We know that  z - score for 98% confidence interval is 2.33 .

Now, 98% confidence interval will be

n-z(SE) and n+z(SE)

Hence, the 98% confidence interval for the students' mean IQ score will be 98.33 - 101.67

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier