What semiannual deposits are needed to accumulate $7,000 in 5 years if the account pays 6% per year, compounded semiannually, assuming that the first deposit is made in 6 months and also assuming that the account already has $1,750 in it today?
Answer(s): A
On the BAII Plus, press 10 N, 6 divide 2 = I/Y, 1750 PV, 7000 +/- FV, CPT PMT. On the HP12C, press 10 n, 6 ENTER 2 divide i, 1750 PV, 7000 CHS FV, PMT. Note that the answer this time is a positive number. This means that the $405.46 is a deposit in addition to the $1,750 original deposit. The $7,000 is entered as a negative number, because the $1,750 and the $405.46 are deposits and the $7,000 is a withdrawal. Make sure the BAII Plus has the value of P/Y set to 1.
A hypothesis test is conducted at the .05 level of significance to test whether or not the population correlation is zero. If the sample consists of 25 observations and the correlation coefficient is 0.60, then what is the computed value of the test statistic?
Answer(s): B
Using the t statistics, t = r* [sq. root of ((n-2)/(1-r_squared))]. t = r*[n-2/(1-r^2)]^0.5. So t = 0.6*[23/0.64]^0.5 = 3.60.
You can enter a derivative contract that will pay $100 at the end of a year if the price of oil exceeds $25 per barrel, or $50 if it is equal to $25 or lower. The probability that oil will exceed $25 by the end of one year is 60%. If interest is 4% for one year, what should the fair price of the contract be?
The expected payoff for the contract is $100 * 0.60 + $50 * 0.40 = $80. At 4% interest, the present value of the expected payoff is $80/1.04 = $76.92. A deviation from this value would represent an example of the investment consequences of inconsistent probabilities.
What monthly payment, beginning next month, would repay a $25,000 car loan over 60 months, assuming your loan has an interest rate of 6.9% per year, compounded monthly?
On the BAII Plus, press 60 N, 6.9 divide 12 = I/Y, 25000 PV, 0 FV, CPT PMT. On the HP12C, press 60 n, 6.9 ENTER 12 divide i, 25000 PV, 0 FV, PMT. Note that the answer will be displayed as a negative number. Make sure the BAII Plus has the value of P/Y set to 1.
A new extended life light bulb has an average service life of 750 hours, with a standard deviation of 50 hours. If the service life of these light bulbs approximates a normal distribution, about what percent of the distribution will be between 600 hours and 900 hours?
Answer(s): E
z = (X-u)/sigma. z1 = (600 - 750)/50 = -3.0. z2 = (900 - 750)/50 = 3.0. from the z tables, z = 3 is 0.4987.Therefore the area between z1 and z2 is 0.4987*2 = 0.9974
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