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Question
In a test on 2,000 electric bulbs, it was found that bulbs of a particular make, was normally distributed with an average life of 2,040 hours and standard deviation of 60 hours. Estimate the number of bulbs likely to burn for more 1,920 hours but less than 2,100 hours
Solution
Let x denote the burning of the bulb follows normal distribution with mean 2,040 and standard deviation 60 hours.
Here m = 2040
σ = 60
N = 2000
The standard normal variate
z = `(x - mu)/sigma`
= `(x - 2040)/60`
P(more 1,920 hours but less than 2,100 hours)
= P(1920 < x < 2100)
When x = 1950
z = `(1920 - 2040)/60`
= `(-120)/60`
= – 2
z = `(2100 - 2040)/60`
= `(-60)/60`
= – 2
∴ P(1920 < x < 2040) = P(– 2 < z < 1)
= P(0 < z < 2) + P(0 < z < 1)
= 0.4772 + 0.3413
= 0.8185
∴ Number of bulbs whose burning time more than 1920 hours but less than 2100 hours)
= 0.8185 × 2000
= 1637
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