Statistics & Applications PYQ — Page 8

The cost of a property appreciates by 10% of the previous month every month. If in end march 2024 it was ₹ 13.31 lakh, when was it ₹ 10 lakh?

A company is shut down due to unavailability of electricity due to non payment of electricity Bill because of some unavoidable circumstances. Under which component of time series does this situation fall?

The effective rate per annum equivalent to a nominal rate of 8% compounded semi-annually is

A machine costing Rs 50,000 has a useful life of 4 years.The estimated scrap value is Rs 10,000 . The rate of depreciation per annum is:

If $x_1, x_2, x_3, ..., x_n$ are n observations in a sample then which of following is/are TRUE? (A) The mean $\bar{x}$ has n degree of freedom (B) The mean $\bar{x}$ has (n-1) degree of freedom (C) The standard deviation of the sample has (n-1) degree of freedom (D) The standard deviation of the sample has n degree of freedom Choose the correct answer from the options given below:

A machine costing ₹ 3,00,000 will have its scrap value of ₹ 50,000. The company at present plans to put ₹ 36,650 per annum at the end of each year in a sinking fund at the rate 5% per annum for the replacement of the machine after its useful life. Suppose the new machine will cost ₹ 4,00,000 at that time, then the useful life (approx.) of the machine is : [Given: $(1.4775)^{1/8} = 1.05$]

Match List-I with List-II | List-I | List-II | |---|---| | (A) A fire in a factory causing production delay for some time is | (I) Secular trend | | (B) Technological progress is | (II) Seasonal trend | | (C) The rise in prices before big festival is example of | (III) Irregular trend | | (D) Rise and fall of share market is | (IV) Cyclic trend | Choose the correct answer from the options given below:

For 95% confidence interval for a population mean reported to be 132 to 142 with standard deviation $\sigma = 17.85$ then the sample size used in this case, is: [Given that: $Z_{0.125} = 1.96$]

A random sample of size 16 has 53 as mean. The sum of the squares of the deviations taken from mean is 150. If the population mean is 56 then the value of t-test statistic is:

The amount of money needed to ensure for a prize of ₹ 5000 at the begining of each year indefinitely if money is worth 5% compounded annually is:

Maneesh took a loan of ₹ 9,00,800 from bank at an interest rate of 6% per annum for 10 years. If she has to pay the loan back with the help of equal monthly installments (EMI). Then, the EMI using reduced balance method is (approx): [Given: $(1.005)^{-120}=0.5496$]

Calculate Three Yearly moving averages for the following data | Year | 2016 | 2017 | 2018 | 2019 | 2020 | 2021 | |---|---|---|---|---|---|---| | Production (thousand tonnes) | 200 | 220 | 231 | 254 | 202 | 243 |

Data available for profit (₹ Thousands) of a company as | Year | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 | |---|---|---|---|---|---|---|---| | Profit(₹ 000) | 114 | 130 | 126 | 144 | 138 | 156 | 164 | Based on the above data using least square method the trend value for the year '2007' is

A small start-up started making wafers and distributing them to the retailers. After a week, average sales per week were found to be 150 packets. So, to increase the sales, a strategy was used to change the packaging and add a chocolate worth Rs. 5 as a free gift with the pack. After this, a sample of 17 shops was taken, which showed that sales went up with mean 165 and a standard deviation of 25. Check whether the strategy was effective @5%, level of significance ? [Given $ t_{16} (0.05) =2.12$]

The average cost function for a commodity is given by $AC = 0.05x^2 - 5x + 1000 + \frac{3000}{x}$ in terms of output x. The fixed cost is

Match **List-I** with **List-II** | List-I | List-II | |---|---| | (A) Perpetuity | (I) A person deposits a fixed amount every year in his bank account to renovate his house after 10 yrs. | | (B) EMI | (II) A person depositis an amount regularly in his bank account and withdraws in case of need. | | (C) Sinking Fund | (III) A fixed amount is debited from the bank account of a person, every month, against a personal loan. | | (D) Saving Account | (IV) A person purchased a house and rents it out. | Choose the **correct** answer from the options given below:

The effective rate, which is equivalent to a nominal rate of 12% compounded semi-annually, is

A company purchased a machine for ₹ 15,00,000 and its effective life is estimated to be 10 years. A sinking fund is created for replacing the machine at the end of its effective life when its scrap value is ₹ 2,42,000. What amount company should provide, at the end of every year out of profits for the sinking fund if it accumulates an interest of 5% per annum? [Given(1.05)¹⁰=1.629]

A man plans to take a housing loan of Rs 99,53,000 from a bank costing 18% per annum compounded monthly. The loan is to be paid back in 30 years in equal monthly installments (EMI). The EMI by reducing balance method is: [Given $(1.015)^{-360} = 0.0047$]

Components of Time Series are (A) Secular Trend Component (B) Seasonal Component (C) Moving Average Component (D) Cyclical Component Choose the **correct** answer from the options given below:

Based on the data available for the production ($y_i$ in thousand tons) of a cloth factory for 7 years ($x_i$) using the method of least squares, the straight line trend is given by $y - a + bx$ with $\sum y_i = 608, \sum x_i = 0, \sum x_iy_i = 116, \sum x_i^2 = 28$. Then, the increase in production per year is:

A startup company invested ₹ 5,00,000 in shares for 4 years. The value of the investment was ₹ 5,50,000 at the end of first year, ₹ 5,25,000 at the end of third year, and on maturity, the final value stood ₹ 6,25,000. The CAGR on the investment will be :- [Given : $(1.25)^{\frac{1}{4}} = 1.06$]

The marks obtained by five students in a test of Applied Mathematics carrying 100 marks are 49, 58, 67, 92, 99. Then the point estimate of the population mean is

Consider the following hypothesis $H_0: \mu = 315$ and $H_a: \mu \neq 315$ A sample of 60 provided a sample mean of 324.6. The standard deviation ($\sigma$) is 14 and level of significance $\alpha = 0.05$. Then the confidence interval is: [Given: $Z_{a/2}{\frac{14}{\sqrt60}} = 3.54$]