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)

my dear correct answer is option number 3