Problem 1.(32 points) A random sample of 100 Uber rides in Chicago has on average 5.4miles per ride. Assume the distribution of the distance of Uber rides to be approximatelynormal with a standard deviation of 1.3 miles.(a) (5 points) Construct a 99% confidence interval for the average distance (in miles) ofan Uber ride. (round to one decimal place)(b) (4 points) What can we assert with 99% confidence about the possible size of our errorif we estimate the average distance of an Uber ride to be 5.4 miles

a) CI = ( 5,1 ; 5,7 )

b) SE =  0,1  

Step-by-step explanation:

a) Sample random     n  = 100

Mean  =  μ = 5,4

Standard deviation   s = 1,3

CI = 99 %    α = 1 %   α = 0,01     α/2 =  0,005

z(c) for  0,005 is from z-table    z(c)  = 2,575

z(c)  =  ( X - μ ) /s/√n          CI  =  μ ± z(c) * s/√n

CI =  5,4 ± 2,575* 1,3/10

CI = 5,4 ± 0,334

CI = ( 5,1 ; 5,7 )

b) SE = Standard deviation / √n

SE = 1,3 /10       SE =  0,1  

We can support that with 99 % of probability our random variable will be in the CI.

(d)

step-by-step explanation:

given: the line in the graph passes through points (5,30) and (6,26).

then, the equation of line is:

now, put y=a(n) and x=n,

hence, option d is correct.