QUESTION: 873

A bar owner in Wisconsin has just won the Illinois Big Lotto jackpot that pays $200 million dollars in 25 equal installments paid annually. The bar owner can either accept the annual payments starting today or receive a lump sum today, using a discount rate of 9% per year. If the bar owner chooses to accept the lump sum, how much will he receive today?

$85.65 million
$91.33 million
$101.31 million
$105.13 million

Answer(s): A

Explanation:

The annual lotto payments equal 200/25 = 8 million dollars per year. Since the payments start today, the cash stream equals a payment of 8 million dollars today and an ordinary annuity with 24 payments. Thus, the present value of the lotto equals 8 + 8/0.09*(1-1/(1.09^24)) = 85.65 million dollars.

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QUESTION: 874

A sample of size 600 is drawn from a population. The sample mean equals 329. The total width of the 99% confidence interval for the population mean is 893. The estimated population variance equals ________.

Answer(s): C

Explanation:

If z is the z-value corresponding to the specified confidence level, the sample mean is M and the standard deviation is D in a sample size N, the confidence interval is specified as [M - z*D/sqrt(N), M + z*D/sqrt(N)].
Thus, the width of the interval equals 2*z*D/sqrt(N). In the present case, for the 99% confidence interval, the normal probability table gives z = 2.58. The sample standard deviation equals sqrt(893) = 29.88. Therefore, the width of the 99% interval is 2*2.58*29.88/sqrt(600) = 6.30.

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QUESTION: 875

The monthly compounded rate is 15% quoted on an annualized basis. The equivalent quarterly compounded rate is:

15.19%
15.05%
15.31%
14.8%

Answer(s): A

Explanation:

To solve such problems, think about investing a dollar for 1 year. The final amount should be the same under both the quotations. Under quarterly compounded rate, r, $1 grows to (1+r/4)^4 in 1 year. Under monthly compounding, it grows to (1+0.15/12)^12 = 1.1608. Since these two should be equal, we get (1+r/4)^4 = 1.1608, giving r = 15.19%. Note that the quarterly compounded rate must be larger than the monthly compounded rate, ruling out 14.8% automatically.

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QUESTION: 876

An accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before they failed is 19.0 hours. The distribution of the lives approximated a normal distribution. The standard deviation of the distribution was 1.2 hours. About 95.44 percent of the batteries failed between what two values?

16.6 and 21.4
14.1 and 22.1
12.2 and 14.2
None of these answers
8.9 and 18.9

Answer(s): A

Explanation:

9544/2 = 0.4772. For an area under the curve of 0.4772, the z value is (+/-)2. We know that z = (x-u)/sigma.
Therefore, +/-2 = (x-19)/1.2. Rearranging we get x = u +/- z*sigma. So x = 19 +/- 2*1.2 = 19+/-2.4. x is 16.6 and 21.4.

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QUESTION: 873

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QUESTION: 874

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QUESTION: 875

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QUESTION: 876

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