Question
Based on past experience, a bank believes that 9% of the people who receive loans will not make payments on time. The bank has recently approved 200 loans.
What assumptions must be true to be able to approximate the sampling distribution with a normal model?
What are the mean and standard deviation of this model?
Mean: μ=0.09
Enter an integer or decimal number, accurate to at least 3 decimal places.
What is the probability that over 10% of these clients will not make timely payments?
What assumptions must be true to be able to approximate the sampling distribution with a normal model?
What are the mean and standard deviation of this model?
Mean: μ=0.09
Enter an integer or decimal number, accurate to at least 3 decimal places.
What is the probability that over 10% of these clients will not make timely payments?
Answers
GPT-4o mini
To approximate the sampling distribution of the proportion of clients who will not make timely payments with a normal model, we need to ensure certain assumptions are met:
### Assumptions:
1. **Independence**: The individuals who receive loans must be independent of one another. This means the payment behavior of one borrower shouldn’t affect another.
2. **Sample Size**: The sample size must be large enough to use the normal approximation. For a proportion to be approximated well by a normal distribution, both \(np\) and \(n(1-p)\) should be greater than 5, where \(n\) is the number of trials (in this case, the number of loans) and \(p\) is the probability of failure (the proportion of clients who do not make timely payments).
In this case, we have:
- \(n = 200\)
- \(p = 0.09\)
- \(np = 200 \times 0.09 = 18 > 5\)
- \(n(1-p) = 200 \times (1 - 0.09) = 200 \times 0.91 = 182 > 5\)
Since both conditions are satisfied, we can use a normal approximation for the sampling distribution.
### Mean and Standard Deviation:
1. **Mean (μ)** of the sampling distribution is given by:
\[
μ = p = 0.09
\]
2. **Standard Deviation (σ)** of the sampling distribution (standard error) is calculated by:
\[
σ = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.09 \times (1 - 0.09)}{200}} = \sqrt{\frac{0.09 \times 0.91}{200}} ≈ \sqrt{\frac{0.0819}{200}} ≈ \sqrt{0.0004095} ≈ 0.0202
\]
Now that we have the mean and standard deviation:
- Mean: μ = 0.09
- Standard Deviation: σ ≈ 0.020 (rounded to 3 decimal places)
### Probability of Over 10% Not Making Timely Payments:
To find the probability that more than 10% of clients will not make timely payments, we are looking for:
\[
P(\hat{p} > 0.10)
\]
Where \(\hat{p}\) is the sample proportion. We first calculate the corresponding z-score.
1. Find the z-score:
\[
z = \frac{\hat{p} - μ}{σ} = \frac{0.10 - 0.09}{0.0202} ≈ \frac{0.01}{0.0202} ≈ 0.495
\]
2. Now, we can use the standard normal distribution to find the probability:
Use a standard normal table or a calculator:
- Find \(P(Z > 0.495)\).
- This is equivalent to \(1 - P(Z \leq 0.495)\).
Using a standard normal distribution table or calculator:
- \(P(Z \leq 0.495) ≈ 0.689\)
Therefore:
\[
P(Z > 0.495) ≈ 1 - 0.689 = 0.311
\]
### Conclusion:
- **Mean**: μ = 0.09
- **Standard Deviation**: σ ≈ 0.020
- **Probability that over 10% of clients will not make timely payments**: \(P(\hat{p} > 0.10) ≈ 0.311\).
So, the final answers are:
- Mean: **0.09**
- Standard Deviation: **0.020**
- Probability: **0.311** (or approximately 31.1%).
### Assumptions:
1. **Independence**: The individuals who receive loans must be independent of one another. This means the payment behavior of one borrower shouldn’t affect another.
2. **Sample Size**: The sample size must be large enough to use the normal approximation. For a proportion to be approximated well by a normal distribution, both \(np\) and \(n(1-p)\) should be greater than 5, where \(n\) is the number of trials (in this case, the number of loans) and \(p\) is the probability of failure (the proportion of clients who do not make timely payments).
In this case, we have:
- \(n = 200\)
- \(p = 0.09\)
- \(np = 200 \times 0.09 = 18 > 5\)
- \(n(1-p) = 200 \times (1 - 0.09) = 200 \times 0.91 = 182 > 5\)
Since both conditions are satisfied, we can use a normal approximation for the sampling distribution.
### Mean and Standard Deviation:
1. **Mean (μ)** of the sampling distribution is given by:
\[
μ = p = 0.09
\]
2. **Standard Deviation (σ)** of the sampling distribution (standard error) is calculated by:
\[
σ = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.09 \times (1 - 0.09)}{200}} = \sqrt{\frac{0.09 \times 0.91}{200}} ≈ \sqrt{\frac{0.0819}{200}} ≈ \sqrt{0.0004095} ≈ 0.0202
\]
Now that we have the mean and standard deviation:
- Mean: μ = 0.09
- Standard Deviation: σ ≈ 0.020 (rounded to 3 decimal places)
### Probability of Over 10% Not Making Timely Payments:
To find the probability that more than 10% of clients will not make timely payments, we are looking for:
\[
P(\hat{p} > 0.10)
\]
Where \(\hat{p}\) is the sample proportion. We first calculate the corresponding z-score.
1. Find the z-score:
\[
z = \frac{\hat{p} - μ}{σ} = \frac{0.10 - 0.09}{0.0202} ≈ \frac{0.01}{0.0202} ≈ 0.495
\]
2. Now, we can use the standard normal distribution to find the probability:
Use a standard normal table or a calculator:
- Find \(P(Z > 0.495)\).
- This is equivalent to \(1 - P(Z \leq 0.495)\).
Using a standard normal distribution table or calculator:
- \(P(Z \leq 0.495) ≈ 0.689\)
Therefore:
\[
P(Z > 0.495) ≈ 1 - 0.689 = 0.311
\]
### Conclusion:
- **Mean**: μ = 0.09
- **Standard Deviation**: σ ≈ 0.020
- **Probability that over 10% of clients will not make timely payments**: \(P(\hat{p} > 0.10) ≈ 0.311\).
So, the final answers are:
- Mean: **0.09**
- Standard Deviation: **0.020**
- Probability: **0.311** (or approximately 31.1%).