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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 number

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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 less than 1,950 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(less than 1950 hours)

P(X < 1950)

When x = 1950

z = `(1950 - 3040)/60`

= `(-90)/60`

= – 1.5

P(X < 1950) = P(Z <  – 1.5) = P(Z > 1.5)

= 0.5 – 0.4332

= 0.068

Numbers of bulbs whose burning time is less than
1950

= 0.0668 × 2000

= 133.6

= 134   ......(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. (ii) | Page 168

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