Purchasing A regional survey found that 70% of all families who indicated an intention to buy a new car bought a new car within 3 months, that 10% of families who did not indicate an intention to buy a new car bought one within 3 months, and that 22% indicated an intention to buy a new car. If a family chosen at random bought a car, find the probability that the family had not previously indicated an intention to buy a car. Harshbarger, Ronald J.. Mathematical Applications for the Management, Life, and Social Sciences (p. 479). Cengage Learning. Kindle Edition.

If a family chosen at random bought a car, we need to find the probability that the family had not previously indicated an intention to buy a car = P(I'|B) = 0.3362

Step-by-step explanation:

Let the event that a family that intends to buy a car be I

Let the event that a family does not intend to buy a car be I'

Let the event that a family buys a car in those 3 months be B

Let the event that a family does not buy a car in those 3 months be B'

Given,

P(B|I) = 0.70

P(B|I') = 0.10

P(I) = 0.22

P(I') = 1 - P(I) = 1 - 0.22 = 0.78

If a family chosen at random bought a car, we need to find the probability that the family had not previously indicated an intention to buy a car = P(I'|B)

The conditional probability P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

So,

P(B|I) = P(B n I) ÷ P(I)

P(B n I) = P(B|I) × P(I) = 0.70 × 0.22 = 0.154

P(B|I') = P(B n I') ÷ P(I')

P(B n I') = P(B|I') × P(I') = 0.10 × 0.78 = 0.078

P(B) = P(B n I) + P(B n I') = 0.154 + 0.078 = 0.232

P(B') = 1 - 0.232 = 0.768

P(I'|B) = P(B n I') ÷ P(B)

= 0.078 ÷ 0.232 = 0.3362

Hope this Helps!!!

nope

step-by-step explanation:

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17