Question

Past participantin a training program designed to upgrade the skills of production. Line supervisors spent an average of 500 hours on the program with standard deviation of 100 hours. Assume a normal distribution.

(i) What is the probability that a participant selected at random will require no less than 500 hours to complete the program ?

(ii) What is the probablity that a participant selected at random will take between 500 and 650 hours to complete the program ?

(iii) What is the probablity that a participant selected at random will take more than 700 hours to complete the program ?

(iv) What is the probablity that a participant selected at random will take between 550 and 650 hours to complete the program ?

(v) What is the probablity that a participant selected at random will require fewer than 580 hours to complete the program ?

(vi) What is the probablity that a participant selected at random will take between 480 and 570 hours to complete the program ?

Answer 1

here we have,

mean = 500

standard deviarion = 100

In normal distribution, Z = (X - mean) / standard deviation

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so,

i)  the probability that a participant selected at random will require no less than 500 hours to complete the program is,

P(X > 500) = P(Z > (500-500)/100) = P(Z > 0) = 1 - P(Z < 0) = 1 - 0.5 = 0.5 (answer)

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ii) the probablity that a participant selected at random will take between 500 and 650 hours to complete the program is,

P(500 < X < 650) = P((500 - 500)/100 < Z < (650-500)/100) = P(0 < Z < 1.5) = P(Z < 1.5) - P(Z < 0) = 0.9332 - 0.5 = 0.4332 (Rounded to 4 decimal places)

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iii) the probablity that a participant selected at random will take more than 700 hours to complete the program is,

P(X > 700) = P(Z > (700-500)/100) = P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228 (rounded to 4 decimal places)

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iv) the probablity that a participant selected at random will take between 550 and 650 hours to complete the program is,

P(550 < X < 650) = P((550-500)/100 < Z < (650-500)/100) = P(0.5 < Z < 1.5) = P(Z < 1.5) - P(Z < 0.5) = 0.9332 - 0.6915 = 0.2417 (Rounded to 4 decimal places)

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v) the probablity that a participant selected at random will require fewer than 580 hours to complete the program is,

P(X < 580) = P(Z < (580-500)/100) = P(Z < 0.8) = 0.7881 (Rounded to 4 decimal places)

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vi) the probablity that a participant selected at random will take between 480 and 570 hours to complete the program is,

P(480 < X < 570) = P((480-500)/100 < Z < (570-500)/100) = P(-0.2 < Z < 0.7) = P(Z < 0.7) - P(Z < -0.2) = 0.7580 - 0.4207 = 0.3373 (Rounded to 4 decimal places)