Noise levels at 5 airports were measured in decibels yielding the following data: 117,118,140,116,119
1. Construct the 90% confidence interval for the mean noise level at such locations. Assume the population is approximately normal.
2. Calculate the sample mean for the given sample data.
3. Find the critical value that should be used in constructing the confidence interval.

1. 112.364<μ,131.636

2. mean = 122

3. critical value = +-2.1318

Step-by-step explanation:

117+118+140+116+119/5

= 610/5

= 122

the sample mean = 122

the sample standard deviation

s² = (117-122)²+(118-122)²+(140-122)²+(116-122²)+(119-122)²/5-1

= 25+16+324+36+9/4

= 410/4

= 102.5

s² = 102.5

s = √102.5

s= 10.12

SE = s/√n

= 10.12/√5

= 4.52

degree of freedom = 5-1 = 4

critical value of t = +-2.1318 using soft ware

confidence interval =

mean +- t(SE)

122-(2.1318)(4.52), 122+(2.1318)(4.52)

= [112.364, 131.636]

112.364<μ,131.636

therefore these are the answers in the ordre it was given in the question

1. 112.364<μ,131.636

2. mean = 122

3. critical value = +-2.1318

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

abde or 1,2,4, 5 is the answer

step-by-step explanation: