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
For question 1, your answers for part (a) are correct:
a) The probability distribution is as follows:
G1: 480/1000 = 0.48
G2: 200/1000 = 0.2
G3: 290/1000 = 0.29
G4: 30/1000 = 0.03
For part (b), however, your calculation is incorrect. To find the probability that a respondent had 6 or fewer months' worth of emergency living expense savings, you need to add the probabilities of the groups G1, G2, and G3, since these groups represent individuals with 6 or fewer months' worth of savings. The correct calculation is:
b) (480 + 200 + 290) / 1000 = 970 / 1000 = 0.97
Therefore, the probability that a respondent had 6 or fewer months' worth of emergency living expense savings is 0.97 (not 0.68).
For question 2, your answers for part (a) are correct:
a) The probability distribution of the random variable x is as follows:
2: 350/1000 = 0.35
3: 200/1000 = 0.2
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
For part (b), your calculation is correct:
b) The probability that a family chosen at random from those surveyed has more than five members is:
0.066 + 0.01 + 0.004 = 0.08
Therefore, the probability is 0.08.
To check your answers for the first question:
a) The probability distribution is a representation of the likelihood of each possible outcome. You have correctly calculated the probabilities for each group as follows:
G1: 480/1000 = 0.48
G2: 200/1000 = 0.2
G3: 290/1000 = 0.29
G4: 30/1000 = 0.03
So, your answer is correct.
b) To find the probability that a respondent had 6 or fewer months' worth of emergency living expense savings, you need to sum the probabilities of groups G1, G2, and G3. You calculated it as:
(200 + 480)/1000 = 680/1000 = 0.68
Your answer is correct.
For the second question:
a) Your calculations for the probabilities of each family size are correct:
Size 2: 350/1000 = 0.35
Size 3: 200/1000 = 0.2
Size 4: 245/1000 = 0.245
Size 5: 125/1000 = 0.125
Size 6: 66/1000 = 0.066
Size 7: 10/1000 = 0.01
Size 8: 4/1000 = 0.004
Your answer is correct.
b) To find the probability that a family chosen at random has more than five members, you need to sum the probabilities of family sizes 6, 7, and 8. You calculated it as:
(66 + 10 + 4)/1000 = 80/1000 = 0.08
Your answer is correct.
Overall, your answers for both questions are correct. Well done!