Suppose the FAA weighed a random sample of 20 airline passengers during the summer and found their weights to have a sample mean of 180 pounds and sample standard deviation of 30 pounds. Assume the weight distribution is approximately normal.

a.) Find a one sided 95% confidence interval with an upper bound for the mean weight of all airline passengers during the summer. Show you work.

b.) Find a 95% prediction interval for the weight of another random selected airline passenger during the summer. Show you work.

Step-by-step explanation:

Given Parameters

Mean, = 180

total samples, n = 20

Standard dev, = 30

= 1 - 0.95 = 0.05 at 95% confidence level

Df = n - 1 = 20 - 1 = 19

Critical Value, , is given by

a).

Confidence Interval, , is given by the formula

b).

Critical Value, , is given by

Confidence Interval, , is given by

    = 180 +/- 14.0403

    = 165.9597 < < 194.0403

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

the answer is ! ?