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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 - Business Mathematics and Statistics

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प्रश्न

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

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उत्तर

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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पाठ 7: Probability Distributions - Exercise 7.3 [पृष्ठ १६८]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
पाठ 7 Probability Distributions
Exercise 7.3 | Q 5. (iii) | पृष्ठ १६८

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