(iii) at least one child is a girl. 46. Driving Habits A marketing firm surveys

Question

(iii) at least one child is a girl. 46. Driving Habits A marketing firm surveys driving habits; the researcher asks the respondents their gender, age group (under 35 years, 35 years or older), and whether thecy prefer driving a car, SUV, or pickup truck. Find a sample space. How many (iv) the first child and the last child are girls. 48. Dividends and the Dow The Dow Jones Industrial Average consists of 30 blue-chip stocks, On April 8, 2010, 18 of these stocks pay annual dividends of more than $1.00 per share. The remaining 12 stocks pay annual dividends of less than $1.00 per share. Construct a probability model that describes the dividends paid by the stocks in the Dow. outcomes are in the sample space? 47 Four-Child Families Families with 4 children are surveyed and the gender of the children according to birth order is recorded. (a) Construct a probability model describing the experiment. Family Income According to the U.S. Census Bureau, the 2008 income of U.S. families is as described in the table. Income Level Number of Families in thousands $25,000 25,000-$49,999 $50,000-$74,999 75,000-$99,999 2$100,000 28,943.7 29,178.1 20,975.4 13,944.5 24,022.1

Explanation / Answer

46. Let G be a random variable denoting the gender.

Let X be a random variable denoting the age group.

let Y be a random variable denoting the type of car.

Sample space of G, S1 = {Male, Female}

Sample space of X, S2 = {under 35 years, 35 years or older}

Sample space of Y, S3 = {Car, SUV, Pickup Truck}

Then the sample space of the event under consideration, S= S1xS2xS3

={(w1,w2,w3): w1 belongs in S1, w2 belongs in S2, w3 beongs in S3

There are 2*2*3= 12 outcomes in the sample space.

47. The gender of each child can be either of the two, Male or Female. Thus there are two possible outcomes corresponding to each child. Thus for 4 children, there are 2^4=16 possible outcomes.

a)Let X be a random variable denoting the gender order.

Sample space, S={MMMM,MMMF,........,FFFF}

Probability of occurence of each outcome is 1/16.

iii) Probability atleast one child is a girl = 1- Probablity none of the child are girl = 1-Probability all the child are bos = 1-(1/16) =15/16

iv) If the first and last child are girls then the 2nd and 3rd child can be of either gender in 2*2 =4 ways.

Thus, probability that the first and last child are girl = 4/16 = 1/4

48) 18 stocks pay dividend of more than $1 per share and 12 stocks pay dividend less than $1 per share.

Let us define X as a random variable denoting the dividends paid by the stocks in the Dow.

Then, sample space, S= {more than $1, less than $1}

P(more than $1) = 18/30

P(less than $1) = 12/30

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