Question

1) (10pts) use of sleepwear while traveling were listed as follows: In USA Today (September 5, 1996), the results of a survey involving the Underwear Nightgown Nothing Paiamas T-shirt Other Total Male 0.220 0.000 0.160 0.102 0.046 0.0840.003 0.612 0.388 Female 0.024 0.182 0.018 0.073 0.088 Total 0.244 0.182 0.178 0.175 0.134 0.087 Suppose a traveler will be selected at random from all travelers responding to the survey What is the probability that the traveler is a male who sleeps wearing a t-shirt? a) (2pts) b) (2pts) What is the probability the traveler is a female? Assuming the traveler is a male, what is the probability the traveler sleeps in pajamas? e) (3pts) What is the probability that a traveler is male if the traveler sleeps in pajamas or a t-shirt. Therefore, we seek to find: d) (3pts) P(male | pajamas U t-shirt).

Answer 1

a) P(traveler is a male who sleeps wearing t shirt)

= 0.046 (Referring to the value from cell Column: ale, Row: T shirt)

b) P(Traveler is female)

= 0.388 (Referring to the value from cell column: Female, Row: Total)

c) P( sleeps in pajamas / the person is male)

= P(sleeps in pajamas and the person is male) / P(the person is male)

= 0.102 / 0.612

0.167

d) P(male / pajamas $\cup$ t shirt)

P (male and pajamas $\cup$ t shirt) / P(pajamas $\cup$ t shirt)

= (0.102 + 0.046) / (0.175 + 0.134)

= 0.479