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MTH302 Business Mathematics and Statistics MCQs

MTH302-Business Mathematics and Statistics study material for students of virtual university of Pakistan. Download Past(old) Papers solved/unsolved, past assignments, quiz, mcqs, lecture notes, video lectures, handouts, books of MTH302-Business Mathematics and Statistics

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16If arrivals follow a Poisson distribution with mean 1.2 arrivals per minute, find the 75th percentile of waiting times (i.e., 75 percent below).
  • 1.155 minutes (69.3 seconds)
  • 0.240 minutes (14.4 seconds)
  • 1.919 minutes (115.1 seconds)
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15Which statement is incorrect?
  • The triangular distribution always has a single mode.
  • The mean of the triangular distribution is (a + b + c)/3.
  • The triangular distribution is always symmetric.
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14A software developer makes 200 phone calls to its current customers. There is an 8 percent chance of reaching a given customer (instead of a busy signal, no answer, or answering machine). The approximate normal probability of reaching at least 15 customers is
  • .4492
  • .5000
  • .6517
  • .4981
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13If freeway speeds are normally distributed with a mean of μ = 70 mph and σ = 7 mph, about what percent of cars will exceed 78 mph?
  • 34.1%
  • 12.7%
  • 15.8%
  • 87.3%
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12The length of brook trout caught in a certain Colorado stream has a mean of 14 inches and a standard deviation of 3 inches. What lower limit should the State Game Commission set on length, if it is desired that 80 per cent of the catch may be kept by fisherpersons?

  • 12.80 inches
  • 11.48 inches
  • 12.4l inches
  • 12.00 inches
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11The length of brook trout caught in a certain Colorado stream has a mean of 14 inches and a standard deviation of 3 inches. The first quartile for the lengths of brook trout would be
  • 16.01 inches.
  • 11.00 inches.
  • 11.98 inches.
  • 8.16 inches.
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10The length of brook trout caught in a certain Colorado stream has a mean of 14 inches and a standard deviation of 3 inches. What proportion of these trout will be between 12 and 18 inches?

  • .6568
  • .6826
  • .2486
  • .4082
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9Assume that X is normally distributed with a mean μ = $64. Given that P(X ≥ $75) = 0.2981, we can calculate that the standard deviation of X is approximately
  • $20.75
  • $13.17
  • $5.83
  • $7.05
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8The variable Z has a standard normal distribution. The probability P(Z ≤ -1.37) is equal to
  • 0.5853
  • 0.9147
  • 0.4147
  • 0.0853
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7The variable Z has a standard normal distribution. The probability P(1.25 ≤ Z ≤ 2.17) is
  • 0.0906
  • 0.9200
  • 0.4700
  • 0.4850
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