C- having at least 1 boy D all boys 2. In January 2015, a sample of 147 students

Question

C- having at least 1 boy D all boys 2. In January 2015, a sample of 147 students from the Cabarrus College student body was taken. Eighty-nine students were female and 58 students were male. Each student was asked whether they were right-or left-handed. The results are shown in the table below. Use these data to answer the questions below. (For parts (a) (h,) give the probabilities first as a fraction and then as a decimal to the thousandths, and write answers as sentences or use descriptive mathematical notation.) -(c), (e), (f), and Left Right Total 82 89 Female Male Total 50 58 15 132 147 Suppose a student is chosen at random from this sample. a. Find the probability that the student is male. b. Find the probability that the student is left-handed. c. Find the probability that he is left-handed, given that the student is male. d. Find the probability that the student is male, given that the student is left-handed. e. For this sample, are the events "being left-handed" and "being male" independent? Explain your answer. f. Find the probability of being left-handed and male. g. Find the probability of being left-handed or male. Show calculations. h. Are the events "Male" and "Left-handed" mutually exclusive? Why or why not?

Explanation / Answer

A) P(male) = 58/147 = 0.395

B) P(left) = 15/147 = 0.102

C) P(left | male) = P(left and male) / P(male) = (8/147)/(58/147) = 8/58 = 0.138

D) P(male | left) = P(male and left) / P(left) = (8/147) /(15/147) = 8/15 = 0.533

E) No they are not independent, because P(male | left) is not equal to P(male)

F) P(left and male) = 8/147 = 0.054

G) P(left or male) = P(left) + P(male) - P(left an male)

= 15/147 + 58/147 - 8/147

= 65/147

= 0.442

H) no they are not mutually exclusive, because P(male and left) is not zero.

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