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Answer:
The sample proportion mean and standard deviation are
[tex]\mu=p=0.13\\\\\sigma=\sqrt{\frac{p(1-p)}{N} }= \sqrt{\frac{0.13*0.87}{500} }=0.015[/tex]
With reasonable assumptions about the sample, all the conditions are met.
The probability that over 16% of these clients will not make timely payments is P=0.023.
Step-by-step explanation:
We have a sample of 500 loans, of which a proportion of p=0.13 is expected to not be paid on time.
The sample proportion mean and standard deviation are
[tex]\mu=p=0.13\\\\\sigma=\sqrt{\frac{p(1-p)}{N} }= \sqrt{\frac{0.13*0.87}{500} }=0.015[/tex]
The assumptions underlying this model are:
- The sample is randomly selected from the population.
- The sampling distribution of p-hat is approximately normal. There are at least 10 successes and 10 failures in the sample.
- Individual observations are independent of each other.
With reasonable assumptions about the sample, all the conditions are met.
What is the probability that over 16% of these clients will not make timely payments?
[tex]z=\frac{x-\mu}{\sigma} =\frac{0.16-0.13}{0.015}=\frac{0.03}{0.015}=2\\\\P(p>0.16)=P(z>2)=0.023[/tex]
The probability that over 16% of these clients will not make timely payments is P=0.023.
The probability that over 16% of the clients will not make the payment is 0.023.
How to calculate the probability?
The sample proportion mean is 0.13. The standard deviation will be:
= (✓0.13 × ✓0.87)/✓500
= 0.015
The probability that over 16% of the clients will not make the payment will be:
= P[Z > (0.16 - 0.13)/0.015]
= 0.023
In conclusion, the probability is 0.023.
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