Question
A survey was conducted in a certain area to determine the distribution of the number of children per family. The results are shown below:
# of Children 0 1 2 3 4 5 6 7 8 9
Frequency 8 9 5 11 3 2 6 4 1 1
Probability 0.16 0.18 0.10 0.22 0.06 0.04 0.12 0.08 0.02 0.02
Show your work to solve the following.
a. Find the probability that a randomly chosen family has at most 4 children.
b. In a random sample of 600 households, do you think it would be likely to have 100 households with only 2 children? Why or why not?
c. If a household is selected at random, predict how many children the household is likely to have.
# of Children 0 1 2 3 4 5 6 7 8 9
Frequency 8 9 5 11 3 2 6 4 1 1
Probability 0.16 0.18 0.10 0.22 0.06 0.04 0.12 0.08 0.02 0.02
Show your work to solve the following.
a. Find the probability that a randomly chosen family has at most 4 children.
b. In a random sample of 600 households, do you think it would be likely to have 100 households with only 2 children? Why or why not?
c. If a household is selected at random, predict how many children the household is likely to have.
Answers
Answered by
GPT-4o mini
To solve the problems presented based on the provided frequency distribution and probabilities of the number of children per family, let’s analyze each part step by step.
### a. Find the probability that a randomly chosen family has at most 4 children.
To find the probability of a family having at most 4 children, we need to sum the probabilities of having 0, 1, 2, 3, and 4 children.
The probabilities are as follows:
- P(0 children) = 0.16
- P(1 child) = 0.18
- P(2 children) = 0.10
- P(3 children) = 0.22
- P(4 children) = 0.06
Now we sum these probabilities:
\[
P(\text{at most 4 children}) = P(0) + P(1) + P(2) + P(3) + P(4)
\]
\[
P(\text{at most 4 children}) = 0.16 + 0.18 + 0.10 + 0.22 + 0.06
\]
\[
P(\text{at most 4 children}) = 0.72
\]
Thus, **the probability that a randomly chosen family has at most 4 children is 0.72.**
### b. In a random sample of 600 households, do you think it would be likely to have 100 households with only 2 children? Why or why not?
To determine if having 100 households with 2 children is likely, we can calculate the expected number of households with 2 children in a sample of 600 households. The probability of a household having 2 children is given as 0.10.
Calculate the expected number of households with 2 children:
\[
E(\text{households with 2 children}) = \text{Total households} \times P(\text{2 children})
\]
\[
E(\text{households with 2 children}) = 600 \times 0.10 = 60
\]
Now, we can observe that the expected number of households with 2 children is 60. Getting exactly 100 households with 2 children would be significantly higher than the expected value.
Next, we assess its likelihood. Using the binomial distribution \(X\) with parameters \(n = 600\) and \(p = 0.10\), we can use the normal approximation for calculations. The mean \(\mu\) and variance \(\sigma^2\) are:
\[
\mu = np = 600 \times 0.10 = 60
\]
\[
\sigma^2 = np(1-p) = 600 \times 0.10 \times 0.90 = 54
\]
\[
\sigma = \sqrt{54} \approx 7.35
\]
We check how many standard deviations away 100 is from the mean:
\[
z = \frac{X - \mu}{\sigma} = \frac{100 - 60}{7.35} \approx 5.44
\]
A \(z\)-score of 5.44 indicates that 100 households with 2 children is extremely far from the expected value, and the probability of such an occurrence is very small (close to 0).
Thus, **it is unlikely to have 100 households with only 2 children in a sample of 600 households.**
### c. If a household is selected at random, predict how many children the household is likely to have.
To predict the number of children a household is likely to have, we calculate the expected value (mean) of the distribution, denoted as \(E(X)\):
\[
E(X) = \sum (x_i \cdot P(x_i))
\]
Using the values given, we can compute it as follows:
\[
E(X) = (0 \cdot 0.16) + (1 \cdot 0.18) + (2 \cdot 0.10) + (3 \cdot 0.22) + (4 \cdot 0.06) + (5 \cdot 0.04) + (6 \cdot 0.12) + (7 \cdot 0.08) + (8 \cdot 0.02) + (9 \cdot 0.02)
\]
Calculating each term:
\[
E(X) = 0 + 0.18 + 0.20 + 0.66 + 0.24 + 0.20 + 0.72 + 0.56 + 0.16 + 0.18
\]
\[
E(X) = 0 + 0.18 + 0.20 + 0.66 + 0.24 + 0.20 + 0.72 + 0.56 + 0.16 + 0.18 = 2.92
\]
Thus, **the expected number of children a household is likely to have is approximately 2.92 children.**
### a. Find the probability that a randomly chosen family has at most 4 children.
To find the probability of a family having at most 4 children, we need to sum the probabilities of having 0, 1, 2, 3, and 4 children.
The probabilities are as follows:
- P(0 children) = 0.16
- P(1 child) = 0.18
- P(2 children) = 0.10
- P(3 children) = 0.22
- P(4 children) = 0.06
Now we sum these probabilities:
\[
P(\text{at most 4 children}) = P(0) + P(1) + P(2) + P(3) + P(4)
\]
\[
P(\text{at most 4 children}) = 0.16 + 0.18 + 0.10 + 0.22 + 0.06
\]
\[
P(\text{at most 4 children}) = 0.72
\]
Thus, **the probability that a randomly chosen family has at most 4 children is 0.72.**
### b. In a random sample of 600 households, do you think it would be likely to have 100 households with only 2 children? Why or why not?
To determine if having 100 households with 2 children is likely, we can calculate the expected number of households with 2 children in a sample of 600 households. The probability of a household having 2 children is given as 0.10.
Calculate the expected number of households with 2 children:
\[
E(\text{households with 2 children}) = \text{Total households} \times P(\text{2 children})
\]
\[
E(\text{households with 2 children}) = 600 \times 0.10 = 60
\]
Now, we can observe that the expected number of households with 2 children is 60. Getting exactly 100 households with 2 children would be significantly higher than the expected value.
Next, we assess its likelihood. Using the binomial distribution \(X\) with parameters \(n = 600\) and \(p = 0.10\), we can use the normal approximation for calculations. The mean \(\mu\) and variance \(\sigma^2\) are:
\[
\mu = np = 600 \times 0.10 = 60
\]
\[
\sigma^2 = np(1-p) = 600 \times 0.10 \times 0.90 = 54
\]
\[
\sigma = \sqrt{54} \approx 7.35
\]
We check how many standard deviations away 100 is from the mean:
\[
z = \frac{X - \mu}{\sigma} = \frac{100 - 60}{7.35} \approx 5.44
\]
A \(z\)-score of 5.44 indicates that 100 households with 2 children is extremely far from the expected value, and the probability of such an occurrence is very small (close to 0).
Thus, **it is unlikely to have 100 households with only 2 children in a sample of 600 households.**
### c. If a household is selected at random, predict how many children the household is likely to have.
To predict the number of children a household is likely to have, we calculate the expected value (mean) of the distribution, denoted as \(E(X)\):
\[
E(X) = \sum (x_i \cdot P(x_i))
\]
Using the values given, we can compute it as follows:
\[
E(X) = (0 \cdot 0.16) + (1 \cdot 0.18) + (2 \cdot 0.10) + (3 \cdot 0.22) + (4 \cdot 0.06) + (5 \cdot 0.04) + (6 \cdot 0.12) + (7 \cdot 0.08) + (8 \cdot 0.02) + (9 \cdot 0.02)
\]
Calculating each term:
\[
E(X) = 0 + 0.18 + 0.20 + 0.66 + 0.24 + 0.20 + 0.72 + 0.56 + 0.16 + 0.18
\]
\[
E(X) = 0 + 0.18 + 0.20 + 0.66 + 0.24 + 0.20 + 0.72 + 0.56 + 0.16 + 0.18 = 2.92
\]
Thus, **the expected number of children a household is likely to have is approximately 2.92 children.**