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Free CFA-Level-I Exam Braindumps (page: 129)

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PDF with 3963 Questions

Samples of the wires coming off the production line were tested for tensile strength. The statistical results (in PSI) were:
Mean = 500
Median = 500
Mode = 500
Standard deviation = 40
Mean deviation = 32
Quartile deviation = 25
Range = 240
Number is sample = 100
The middle 95 percent of the wires tested between approximately what two values?

  1. 450 and 550
  2. 420 and 580
  3. 380 and 620
  4. 460 and 540
  5. None of these answers

Answer(s): B

Explanation:

95% of the observations lie between plus and minus two standard deviations from the mean, so it is 500 +/- 2 (40).



Theta Corp.'s stock is expected to appreciate 6% per year. If no dividends are paid out over the next 5 year and the current stock price is $28, what's the expected price in 5 years?

  1. $39.6
  2. $37.5
  3. $41.2
  4. $34.9

Answer(s): B

Explanation:

The expected price in 5 years is 28*(1.06^5) = 37.5



The lowest significance level at which the null hypothesis can be rejected is called the:

  1. F-statistic of the test.
  2. t-value of the test.
  3. p-value of the test.
  4. critical value of the test.
  5. none of these answers.

Answer(s): C

Explanation:

The lowest significance level at which the null hypothesis can be rejected is called the p-value of the test. Thus, if the p-value is less than the significance level, the null hypothesis can be rejected at that significance level.



The sponsors of a well-known charity came up with a unique idea to attract wealthy patrons to the $500 a plate dinner. After the dinner, it was announced that each patron attending could buy a set of 20 tickets for the gaming tables. The chance of winning a prize for each of the 20 plays is 50-50. If you bought a set of 20 tickets, what is the chance that you will win 15 or more prizes?

  1. 0.021
  2. 0.250
  3. 0.006
  4. None of these answers
  5. 0.750

Answer(s): A

Explanation:

This is a binomial probability. The probability of getting r successes out of n trials where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p) is given by: n!(p^r)[q^(n-r)]/r!(n-r)!.
Here n = 20, p = 0.5 and q = 0.5 and r = 15,16,17,18,19,20. Therefore we have P(15) = 20!(0.5^15)(0.5^5)/15!5! = 0.0148
P(16) = 20!(0.5^16)(0.5^4)/16!4! = 0.0046
P(17) = 20!(0.5^17)(0.5^3)/17!3! = 0.0011
P(18) = 20!(0.5^18)(0.5^2)/18!2! = 0.0002
P(19) = 20!(0.5^19)(0.5^1)/19!1! = 0.00002
P(20) = 20!(0.5^20)(0.5^0)/20!0! = 0.000001
The sum adds up to 0.207.






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