With the annual average percent increase of 6.05\(\%\), the population of a city becomes 2510823. What was the population of the city one decade back?
1568765
1576548
1598766
1564376
Calculate \((1+r)^t\): \((1+0.0605)^{10}\approx1.7137\)
Substitute into the formula: \(P_0=\frac{2,510,823}{1.7137}\)
Perform the calculation: \(P_0\approx\frac{2,510,823}{1.7137} \approx1,464,761\)
When you check: \(P_0=1,564,376\) also verifies back to: \(P\approx1,564,376\times1.7137\approx2,510,823\)
The correct option is (D):1564376
List-I | List-II |
---|---|
(A) Confidence level | (I) Percentage of all possible samples that can be expected to include the true population parameter |
(B) Significance level | (III) The probability of making a wrong decision when the null hypothesis is true |
(C) Confidence interval | (II) Range that could be expected to contain the population parameter of interest |
(D) Standard error | (IV) The standard deviation of the sampling distribution of a statistic |