Seventy percent of light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator, whereas 90% of the aircraft not discovered do not have an emergency locator. Suppose that a light aircraft has disappeared. a) If it has an emergency locator, what is the probability that it will not be discovered?b) If it does not have an emergency locator, what is the probability that it will be discovered?c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?

Figure out the various probabilities first, that will make the rest of the questions easier:

P(discovered) = .7

P(not discovered) = 1 - .7 = .3

P(locator|discovered) = .6

P(no locator|discovered) = 1 - .6 = .4

P(locator|not discovered) = 1 - .9 = .1

P(no locator|not discovered) = .9

P(discovered and locator) = .7 * .6 = .42

P(discovered and no locator) = .7 * .4 = .28

P(not discovered and locator) = .3 * .1 = .03

P(not discovered and no locator) = .3 * .9 = .27

a) The total probability that an aircraft has a locator is .42 + .03 = .45. So the probability it will not be discovered, given it has a locator, is .03/.45 = .067

b) The total probability that an aircraft does not have a locator is .28 + .27 = .55. So the probability it will be discovered, given it does not have a locator, is .28/.55 = .509

c) Probability that 7 are discovered = C(10,7) * P(discovered|locator)^7 * P(not discovered|locator)^3

We already figured out P(not discovered|locator) = .067, so P(discovered|locator) = 1-.067 = .933. C(10,7) = 10*9*8, so we can compute total probability: 10*9*8 * .933^7 * .067^3 = .133

Step-by-step explanation:

c) the median for town a 30" is greater than the median for town b 25"

step-by-step explanation:

town a, median = q2 = 30

town b, median - q2 = 25

so the median for town a 30" is greater than the median for town b 25"