Answer to Question #198561 in Statistics and Probability for fahmida

Question

4) To investigate how often families eat at home, Harris Interactive surveyed 500 adults living with children under the age of 18. The survey results are shown in the following table.

Number of Family Meals per Week - 0,1,2,3,4,5,6,7 or more.

Number of Survey Responses - 20,30,40,112,66,52,56,124.

For a randomly selected family with children under the age of 18, compute the following.

1. The probability of the family eats no meals at home during the week.
2. The probability of the family eats at least five meals at home during the week.
3. The probability of the family eats three or fewer meals at home during the week.

Consider tossing two coins.

1. Develop the sample space.

Answer 1

1.)The probability of the family eats no meals at home during the week $= \dfrac{11}{496}$

2.)The probability of the family eats at least five meals at home during the week.

$= \dfrac{119+114+139}{496} = \dfrac{372}{496}$

3.)The probability of the family eats three or fewer meals at home during the week.

$= \dfrac{11+11+30+36}{496} = \dfrac{88}{496}$

1.) For tossing two coins, the corresponding sample space would be {(head,head), (head,tail), (tail,head), (tail,tail)}, commonly written {HH, HT, TH, TT}.