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3 years ago

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A person borrowef Rs 16000 from a bank at 12.5% pet annum simple intrest and lent the whole amount to shopkeeper at the sam rate

of compound interest.How much will he gain in 2 years?

1 answer:

r-ruslan [8.4K]3 years ago

8 0

Answer:

The amount he gains in 2 years is Rs. 250

Step-by-step explanation:

The parameters of the amount borrowed from the bank are;

The amount borrowed, P = Rs. 16,000

The interest rate of the amount borrowed, R = 12.5%

The number of years, T = 2 years

The simple interest, I = (P × R × T)/100 = (Rs. 16,000 × 12.5 × 2)/100 = Rs. 4,000

The total amount the person is to pay bank to the bank = P + I = Rs. 16,000 + Rs. 4,000 = Rs. 20,000

The parameters of the amount lent to the shopkeeper are;

The amount lent, P = Rs. 16,000

The compound interest rate of the amount lent, R = 12.5%

The number of years, T = 2 years

The amount he receives from the shopkeeper after 2 years, A = P·(1 + R/n)^(n × T) = Rs. 16,000 × (1 + 0.125/1)^(1 × 2) = Rs. 20,250

The amount he gains in 2 years = The amount he receives from the shop keeper - The amount he gives to the bank = Rs. 20,250 - Rs. 20,000 = Rs. 250

The amount he gains in 2 years = Rs. 250.

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Complete the tasks and answer the questions. Use the 2SD method to estimate the true proportion of the population of your city t

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Complete Question

The complete question is shown on the first uploaded image

Answer:

Using 2SD Method

The confidence interval is

The interval notation : $(0.108 , 0.212 )$

The interval notation and the sample proportion ± the margin of error

notation: $0.16 \pm 0.052$

The interpretation

There is 95% confidence that the true proportion of those that dislike cilantro lie within the upper(0.212) and the lower(0.108) limit of the calculate confidence interval

Using theory-based method

The 95% confidence interval is

The interval notation : $(0.109 , 0.211 )$

The interval notation and the sample proportion ± the margin of error

notation: $0.16 \pm 0.051$

The interpretation

There is 95% confidence that the true proportion of those that dislike cilantro lie within the upper(0.211) and the lower(0.109) limit of the calculate confidence interval

Using theory-based method

The 88% confidence interval is

The interval notation : $(0.120 , 0.200 )$

The interval notation and the sample proportion ± the margin of

error notation: $0.16 \pm 0.040$

The interpretation is

There is 88% confidence that the true proportion of those that dislike cilantro lie within the upper(0.200 ) and the lower(0.120) limit of the calculate confidence interval.

Step-by-step explanation:

From the question we are told that

The number sample size is n = 200

The number of people that dislike cilantro is k = 32

Generally the sample proportion is mathematically represented as

$\r p = \frac{k}{N}$

=> $\r p = \frac{32}{200}$

=> $\r p = 0.16$

Generally 2SD confidence interval is mathematically represented as

$\r p \pm 2\sqrt{\frac{\r p(1 - \r p)}{n} }$

substituting value

$0.16 \pm 2\sqrt{\frac{0.16(1 - 0.16)}{200} }$

=> $0.16 \pm 0.052$

This can also be represented as

$(0.16 - 0.052 , 0.06 + 0.052)$

=> $(0.108 , 0.212 )$

Generally this confidence interval can be interpreted as

There is 95% confidence that the true proportion of those that dislike cilantro lie within the upper and the lower limit of the calculate confidence interval

Using theory-based method estimate 95% confidence interval

Generally from the question the confidence level is 95% hence the level of significance is calculated as

$\alpha = (100 - 95)\%$

=> $\alpha = 0.05$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the confidence level using theory base method is

$0.16 \pm 1.96 \sqrt{\frac{0.16 (1- 0.16)}{200} }$

=> $0.16 \pm 0.051$

This can also be represented as

$(0.16 - 0.051 , 0.06 + 0.051)$

=> $(0.109 , 0.211 )$

Generally this confidence interval can be interpreted as

There is 95% confidence that the true proportion of those that dislike cilantro lie within the upper(0.211) and the lower(0.109) limit of the calculate confidence interval

Using theory-based method estimate 88% confidence interval

Generally from the question the confidence level is 95% hence the level of significance is calculated as

$\alpha = (100 - 88)\%$

=> $\alpha = 0.12$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.12 }{2} } = 1.55$

Generally the confidence level using theory base method is

$0.16 \pm 1.55 \sqrt{\frac{0.16 (1- 0.16)}{200} }$

=> $0.16 \pm 0.040$

This can also be represented as

$(0.16 - 0.040 , 0.06 + 0.040)$

=> $(0.120 , 0.200 )$

Generally this confidence interval can be interpreted as

There is 88% confidence that the true proportion of those that dislike cilantro lie within the upper(0.200 ) and the lower(0.120) limit of the calculate confidence interval

3 0

4 years ago