Suppose that many large samples are taken from a population and that the
sample proportions are normally distributed, with a mean of 0.22 and a
standard error of 0.029. What is the 68% confidence interval?
O
O
O
O
A. (0.162 to 0.278)
B. (0.104 to 0.336)
C. (0.133 to 0.307)
D. (0.191 to 0.249)

h/o/e/s mad

step-by-step explanation:

si si si

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %