Question

2. In households with children, some of the school age children ate school lunches and others did not. Hence, we have another random phenomenon. Define a new random variable v as follows: O, if none of the children in household ate school lunch 1, if at least one child in household ate school lunch a. Table 20.8 contains data collected from a U.S. government survey on random variable v. Calculate the proportions for the outcomes of v and enter them as estimates in the Probability column. Round probabilities to three decimals. Unit 20: Random Variables Student Guide | Page 13 V Frequency Probability 0 32,491 1 74,690 Total Table 20.8. Data on random variable v. b. Draw a diagram that represents all households broken down by whether or not there are children in the household, and then broken down further by whether or not at least one child in the household ate school lunch. Determine the proportion of all households that had at least one child who ate school lunch.

Answer 1

Question (a)

v	Frequency	Probability
0	32.491	0.303
1	74,690	0.697
	$\sum$frequency = 107181	$\sum$Probability = 1

The probabilities of the random varibale v are obtained by dividing respective fequency by the total frequency

Probabity of v=0 is  32491/107181 = 0.303

Probabity of v=1 is 74690/107181 = 0.697

Proportion of outcome v = 0 is 30.3%

Poroportion of outvome v = 1 is 69.7%

Question (b)

The tree digaram is as follows

Households Households with children Households without children 30.3% 69.7% None of the children in the household ate school lunch At least one of the children in the household ate school lunch

So proportion of househilds that had atleast one child ate all school luch is 69.7% or 0.697 as calculated in the table in Question (a)