Section 5.2: Normal Distributions: FindProbabilities 2. The diameters of a woode

Question

Section 5.2: Normal Distributions: FindProbabilities

2. The diameters of a woodendowel produced by a new machine are normally distributed with amean of 0.55 inches and a standard deviation of 0.01 inches. Whatpercent of the dowels will have a diameter greater than 0.57?

(5 points)

3. A loan officer ratesapplicants for credit. Ratings are normally distributed. The meanis 240 and the standard deviation is 50. Find the probability thatan applicant will have a rating greater than 260.

          (5points)

Explanation / Answer

2) The empirical rule states:
Mean +/- the standard deviation will contain approximately 68% ofthe measurements.
Mean +/- 2* standard deviation will contain approx 95% of themeasurements
Mean +/- 3* standard deviation will contain approx 99.7% of themeasurements

In this case, we are looking at the mean plus 2 standarddeviations.
We also want the percentage of ALL the measurements less than .57,so this is the same as 1 - percentage greater than .57.

The upper tail contails 2.5% of our measurements, so the answeris
1-.025 = .975 or 97.5% or measurements 3)

• • Mean = 240 and SD = 60
    
    Then SE = 60/sqrt(36) = 10
    
    P(230 < X < 260) = P( (230-240)/10 < Z <(260-240)/10)
    = P(-1 < Z < +2)
    = P(-1 < Z) - P(+2 < Z)
    = 0.84134 - 0.02275
    = 0.81859
  • 19.7%
    
    normalcdf(lowerbound, upperbound, mean, standard deviation)
    
    or
    
    normalcdf(230, 260, 240, 60) = .1967, or 19.7%

• Mean = 240 and SD = 60
  
  Then SE = 60/sqrt(36) = 10
  
  P(230 < X < 260) = P( (230-240)/10 < Z <(260-240)/10)
  = P(-1 < Z < +2)
  = P(-1 < Z) - P(+2 < Z)
  = 0.84134 - 0.02275
  = 0.81859
• 19.7%
  
  normalcdf(lowerbound, upperbound, mean, standard deviation)
  
  or
  
  normalcdf(230, 260, 240, 60) = .1967, or 19.7%

• Mean = 240 and SD = 60
  
  Then SE = 60/sqrt(36) = 10
  
  P(230 < X < 260) = P( (230-240)/10 < Z <(260-240)/10)
  = P(-1 < Z < +2)
  = P(-1 < Z) - P(+2 < Z)
  = 0.84134 - 0.02275
  = 0.81859
• 19.7%
  
  normalcdf(lowerbound, upperbound, mean, standard deviation)
  
  or
  
  normalcdf(230, 260, 240, 60) = .1967, or 19.7%

Mean = 240 and SD = 60

Then SE = 60/sqrt(36) = 10

P(230 < X < 260) = P( (230-240)/10 < Z <(260-240)/10)
= P(-1 < Z < +2)
= P(-1 < Z) - P(+2 < Z)
= 0.84134 - 0.02275
= 0.81859 Mean = 240 and SD = 60

Then SE = 60/sqrt(36) = 10

P(230 < X < 260) = P( (230-240)/10 < Z <(260-240)/10)
= P(-1 < Z < +2)
= P(-1 < Z) - P(+2 < Z)
= 0.84134 - 0.02275
= 0.81859 19.7%

normalcdf(lowerbound, upperbound, mean, standard deviation)

or

normalcdf(230, 260, 240, 60) = .1967, or 19.7% 19.7%

normalcdf(lowerbound, upperbound, mean, standard deviation)

or

normalcdf(230, 260, 240, 60) = .1967, or 19.7% 19.7%

normalcdf(lowerbound, upperbound, mean, standard deviation)

or

normalcdf(230, 260, 240, 60) = .1967, or 19.7%

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