According to the MIT Airline Data Project, American Airlines controlled 15.5% of the domestic market during a recent year. A random sample of 125 domestic passengers that year was selected. Using the normal approximation to the binomial distribution, what is the probability that 10 or fewer passengers from this sample were on American Airline flights

0.0143

Step-by-step explanation:

In this question, we are asked to use the binomial distribution to calculate the probability that 10 or fewer passengers from a sample of MIT data project sample were on American airline flights.

We proceed as follows;

The probability that a passenger was an American flight is 15.5%= 15.55/100 = 0.155

Let’s call this probability p

The probability that he/she isn’t on the flight, let’s call this q

q =1 - p= 0.845

Sample size, n = 155

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 125 x 0.155

= 19.375

Standard deviation = √npq

= √ (125 x 0.155x 0.845)

= 4.0462

P(10 or fewer passengers were on American Airline flights) = P(X \leq 10)

= P(Z < (10.5 - 19.375)/4.0462)

= P(Z < -2.19)

= 0.0143

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

step-by-step explanation:

sry but any answer choices