is my answer to a correct? if not please show me what i did wrong. also please s

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is my answer to a correct? if not please show me what i did wrong. also please show me how to get part b. i dont understandconfidence intervals.
A university has 30000 registered students. As partof a survey, 900 of these students were chosen at random. Theaverage age of the sample students turns out to be 22.3 andthe sd is 4.5 years.
a)The average age of all 3000 students is estimated as ___. This estimate is likely to be off by ___ years or so. first blank: 22.3 second blank: 4.5
b) Find a 95% confidence interval for the average age of all30000 registered students.
A university has 30000 registered students. As partof a survey, 900 of these students were chosen at random. Theaverage age of the sample students turns out to be 22.3 andthe sd is 4.5 years.
a)The average age of all 3000 students is estimated as ___. This estimate is likely to be off by ___ years or so. first blank: 22.3 second blank: 4.5
b) Find a 95% confidence interval for the average age of all30000 registered students.
a)The average age of all 3000 students is estimated as ___. This estimate is likely to be off by ___ years or so. first blank: 22.3 second blank: 4.5
b) Find a 95% confidence interval for the average age of all30000 registered students.

Explanation / Answer

sample mean, x = 22.3 standard deviation, s = 4.5 sample size, n = 900 standard error, e = s/n = 4.5/900 =0.15 a)The average age of all 3000 students is estimated as ___. This estimate is likely to be off by ___ years or so. first blank: 22.3 second blank: 4.5
b) at 95% confidence for a two tail distribution, z =1.96       (from z table) 95% confidence interval for the average age of all 30000registered students = x ± z*e = 22.3 ± (1.96*0.15) = ( 22.006, 22.594)
a)The average age of all 3000 students is estimated as ___. This estimate is likely to be off by ___ years or so. first blank: 22.3 second blank: 4.5
b) at 95% confidence for a two tail distribution, z =1.96       (from z table) 95% confidence interval for the average age of all 30000registered students = x ± z*e = 22.3 ± (1.96*0.15) = ( 22.006, 22.594)

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