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Free CFA-Level-I Exam Braindumps (page: 273)

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PDF with 3963 Questions

The seasonal output of a new experimental strain of pepper plants was carefully weighed. The mean weight per plant is 15.0 pounds, and the standard deviation of the normally distributed weights is 1.75 pounds. Of the 200 plants in the experiment, how many produced peppers weighing between 13 and 16 pounds?

  1. 100
  2. None of these answers
  3. 118
  4. 53
  5. 197

Answer(s): C

Explanation:

z = (x-u)/sigma. z1 = 13 - 15/1.75 = -1.1429 and z2 = 16 - 15/1.75 = 0.5714. The respective areas for those z- values are 0.3729 and 0.2157. Since they are on opposite sides of the mean, we add them up to find the area in between which is 0.5886. Therefore, 0.5886*200 = 118.



If you deposit $1,500 into an account paying 6% per year, compounded quarterly, how much is in the account after 60 months?

  1. $2,224.46
  2. $2,117.54
  3. $2,020.28
  4. $1,702.48
  5. $1,751.59

Answer(s): C

Explanation:

There are 20 quarters in 60 months (60 divide 3 =). On the BAII Plus, press 20 N, 6 divide 4 = I/Y, 1500 PV, 0 PMT, CPT FV. On the HP12C, press 20 n, 6 ENTER 4 divide i, 1500 PV, 0 PMT, FV. 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.



BeigeBox, a computer manufacturer, can construct computer systems with any one of 4 processors, 2 memory configurations, 3 hard drives, 2 monitors, 2 keyboards, and 2 CD drives. How many unique configurations of computer systems can BeigeBox construct?

  1. 192.
  2. 96.
  3. 384.
  4. 64.

Answer(s): A

Explanation:

The multiplication rule of counting states that the number of combinations available when there are n_1 options in one aspect, n_2 in another, and so on, up to n_k, is n_1
* n_2 * ... * n_k. In this case, the number of combinations is 4*2*3*2*2*2 = 192.



In a beer taste study involving the ExtraFerm lager beer, the following responses were recorded:
Response Frequency
Great taste 103
Okay taste 46
Bad taste 64
The probability that a randomly selected beer drinker will NOT find ExtraFerm to be bad tasting is calculated from these data to be:

  1. 0.22
  2. 0.30
  3. 0.70
  4. 0.78

Answer(s): C

Explanation:

The probability that a randomly selected beer drinker will find ExtraFerm to be bad tasting is estimated from the above data as 64/(64 + 46 + 103) = 0.3. Therefore, the probability that a randomly selected beer drinker will NOT find ExtraFerm to be bad tasting equals 1 - 0.3 = 0.7. We could also have calculated this probability as (103 + 46)/(64 + 46 + 103) = 0.7 Note that given a frequency distribution, the estimated probability of a particular class/event occurring equals the relative frequency of that class.






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