A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the college are normally distributed with a standard deviation of 1.8 years. The 98% confidence interval for the average age of all students at this college is .

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = ~ N(0,1)

where, = sample average age = 25 years

            = population standard deviation = 1.8 years

            n = sample of students = 36

            = population average age

So, 98% confidence interval for the average age, is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < < 2.3263) = 0.98

P( < < ) = 0.98

P( < < ) = 0.98

98% confidence interval for = [ , ]

                                                  = [ , ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

16

step-by-step explanation:

2 x 40 = 80

80/5

= 16