Concept explainers
a
Interpretation:
Proportion of bulb lasting more than 2 years.
Concept Introduction:
Mean is the average value of the data given. It is usually the middle value which represents the whole data. It is calculated by summing up all values divided by umber of values.
a
Answer to Problem 42AP
The probability of bulbs lasting more than two years is 0.48%.
Explanation of Solution
Given information:
Time Period = 2 year
Mean = 2.670
The proportion of bulbs lasting more than two years using the following formula:
Here, t is 2 years and
Thus, the probability of bulbs lasting more than two years is 0.48%.
b
Interpretation:
Probability that a bulb chosen fails in first-three months of operation.
Concept Introduction:
Mean is the average value of the data given. It is usually the middle value which represents the whole data. It is calculated by summing up all values divided by umber of values.
b
Answer to Problem 42AP
Probability that a bulb chosen at random fails is 48.69%.
Explanation of Solution
Given information:
Time = 3 months
Mean = 2.670
The probability that a bulb chosen at random fails using the following formula:
Here, t is 3/12 years and
Thus, the probability that a bulb chosen at random fails is 48.69%.
c
Interpretation:
Probability that a bulb lasted for 10 years fails in next three months.
Concept Introduction:
Mean is the average value of the data given. It is usually the middle value which represents the whole data. It is calculated by summing up all values divided by umber of values.
c
Answer to Problem 42AP
The probability of bulbs that has lasted for 10 years fails in the next three months of operation is1.9102E+13%.
Explanation of Solution
Given information:
Time = 10 years
Mean = 2.670
The probability that a bulb that has lasted for 10 years fails in the next three months of operation using the following formula:
Here, t is 10 years and s is 3/12. Substitute the values in the equation and calculate the proportion of bulb that has lasted for 10 years fails in the next three months of operation as shown below:
Thus, P (T < s) is 0.486.
Thus, P (T > 10) is
Substitute the values in the equation (1) and calculate as shown below:
Thus, the probability of bulbs that has lasted for 10 years fails in the next three months of operation is 1.9102E+13%.
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Chapter 13 Solutions
Production and Operations Analysis, Seventh Edition
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