# 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 _____.

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Answer: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. = [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1) where, [tex]\bar X[/tex] = sample average age = 25 years [tex]\sigma[/tex] = population standard deviation = 1.8 years n = sample of students = 36 [tex]\mu[/tex] = population average ageSo, 98% confidence interval for the average age, [tex]\mu[/tex] is ;P(-2.3263 < N(0,1) < 2.3263) = 0.98P(-2.3263 < [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 2.3263) = 0.98P( [tex]-2.3263 \times {\frac{\sigma}{\sqrt{n} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.3263 \times {\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.98P( [tex]\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.9898% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }[/tex] , [tex]\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }[/tex] ] = [ [tex]25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }[/tex] , [tex]25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }[/tex] ] = [24.302 , 25.698]Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].