The lifetime (in hours) of a 60-watt light bulb is a random variable that has a normal distribution with σ = 30 hours. a random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038. construct a 92% confidence interval estimate for the mean lifetime μ. if it were desired to cut the confidence interval to half its length while keeping the same 92% level, what size sample would be required to achieve this?

a) 92% Confidence interval: (1027.5,1048.5)

b) Sample size = 100

Step-by-step explanation:

We are given the following in the question:

Sample mean, = 1038

Sample size, n = 25

Alpha, α = 0.08

Population standard deviation, σ = 30

a) 92% Confidence interval:

Putting the values, we get,

b) In order to reduce the confidence interval by half, we have to quadruple the sample size.

Thus,

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841