Answer to Question #198578 in Statistics and Probability for fahmida

Question

11) To investigate how often family members eat at home, Maya 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.

a) The probability of the family eats no meals at home during the week.

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

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

Answer 1

Solution:

Total number of adults surveyed = 500

Let $X$ be the random variable denoting no. of family meals per week.

(a) $P(X=0) = \dfrac{20}{500}=0.04$

(b) $P(X\ge5) = P(X=5)+P(X=6)+P(X\ge7)$

$=\dfrac{52+56+124}{500}=\dfrac{232}{500}=0.464$

(c) $P(X\le3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$=\dfrac{20+30+40+112}{500}=0.404$