Suppose that a random sample of eighteen recently sold houses in a certain city has a mean sales price of , with a standard deviation of . Under the assumption that house prices are normally distributed, find a confidence interval for the mean sales price of all houses in this community. Then find the lower limit and upper limit of the confidence interval.

This question is incomplete, the complete question is;

Suppose that a random sample of eighteen recently sold houses in a certain city has a mean sales price of $280,000, with a standard deviation of $10,000. Under the assumption that house prices are normally distributed, find a 95% confidence interval for the mean sales price of all houses in this community. Then find the lower limit and upper limit of the confidence interval.

At 95%, confidence interval is ( 275026.7, 284973.3 )

Hence;

Lower limit = 275026.7

Upper limit = 284973.3

Step-by-step explanation:

 Given that;

mean x = $280,000

standard deviation σ = $10,000

sample size n = 18

degree of freedom df = n - 1 = 18 - 1 = 17

∝ = 1 - 95% = 1 - 0.95 = 0.05  

so

Critical t value = tinv( 0.05, 17 ) = 2.11

now, at 95% confidence interval for mean will be;

⇒ x ± [ Critical t value × ( σ/√n) ]

so we substitute

⇒ 280,000 ± [  2.11 × ( 10000/√18) ]

⇒ 280,000 ± [  2.11 × 2357.0226 ]

⇒ 280,000 ± 4973.3

⇒ 280000 - 4973.3, 280000 + 4973.3

At 95%, confidence interval is ( 275026.7, 284973.3 )

Hence;

Lower limit = 275026.7

Upper limit = 284973.3

the total estimated nappy use.=10,500,000

step-by-step explanation:

given that on an average the average baby uses 3528 napkins.

in a particular year 2014, the number of births were 299697 in new south wales.

rounding off the number of births to the nearest hundred thousand we have

no of birtsh = 300000

also rounding off the average usage to nearst thousand we get

average use of a baby = 3500

hence total number of nappy estimated to be used = 3500x300000

= 10,500,000

answer is 10,500,000 nappies