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Tamil Nadu Board of Secondary EducationHSC Commerce Class 12

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

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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 than 2,150 hours

Sum
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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(morethan 2,150 hours)

P(X > 2150)

When x = 2150

z =  `(2150 - 2040)/60`

= `110/60`

= 1.833

P(X > 2150) = P(Z > 1.833)

= P(0 < z < `oo`) – P(0 < z < 1.833)

= 0.5 – 0.4664

= 0.0336

∴ Number of bulbs whose burning time is more than 2150 hours

= 0.0336 × 2000

= 67.2

= 67  .......(approximately)

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Chapter 7: Probability Distributions - Exercise 7.3 [Page 168]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 12 TN Board
Chapter 7 Probability Distributions
Exercise 7.3 | Q 5. (i) | Page 168

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