I only need someone to check my answers for two questions.
1) In a survey of 1000 people about emergency living expense savings .... the following results were retained (basically the number of months the person could live on savings):
Group 1 2 3 4
Time (months) Less than 3 3-6 More than 6 No answer
Number 480 200 290 30
Let x denote the random variable that takes on the values 1,2,3,and 4, corresponding to the groups given in the table.
a) Find the probability distribution
b) If a respondent in the survey is chosen at random, what is the probability that he or she had 6 or fewer months' worth of emergency living expense savings?
My answers: (G stands for group)
a) G1 :480/1000 =.48 G2: 200/1000 =.2, G3: 290/1000 =.29
G4: 30/1000=.03
b) 200+480/1000= 680/1000 =.68
2)... survey of 1000 families to determine the distribution of families by size. The results follow:
Family size: 2 3 4 5 6 7 8
Frequency of occurrence: 350 200 245 125 66 10 4
a) find the probability distribution of the random variable x, where x denotes the number of people in a randomly chosen family.
b) Find the probability that a family chosen at random from those surveyed has more than five members.
a) 2: 350/1000 =.35, 3: 200/1000=.2, 4:245/1000=0.245 5: 125/1000 = 0.125, 6: 66/1000 =0.066, 7: 10/1000= 0.01, 8: 4/1000=0.004
b) add family sizes 6,7,8 (66+10+4)
= 80/1000= 0.08
5 answers