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Which one of mean or standard deviation can used to solve the following problem?


A light bulb is considered defective if it lasts less than 400 hours. The following claim is made:

'Brand A light bulbs are more likely to be defective than Brand B light bulbs.'

Is the claim correct?

$$ \begin{array}{c|lc} & \text{} & \text{Mean} & \text{Standard deviation}\\ & Brand A & 450 & 25 \\ & Brand B & 500 & 50 \\ \end{array} $$


My guess is that the claim is incorrect. The reason is that Brand B standard deviation is higher. This shows that although Brand B has a higher mean but the data is more distributed. However I cannot proof my guess. Is there a way that I can find out the number of bulbs that had a higher lifetime of 400, so I can make a comparison?

bman
  • 675

1 Answers1

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Assuming that "lamp lasts" distributed normally, we can find, that Z-score for brands A and B are

Z(A) = (450 - 400) / 25 = 2 sigma

Z(B) = (500 - 400) / 50 = 2 sigma

so we can conclude, that probabilities of being defective are the same for both A and B brands; according one tail normal distribution it equals to

P = 0.0228...

  • Thank you for the answer. I just didn't understand how you find P = 0.0228... ? @Dmitry Bychenko – bman Jun 23 '13 at 10:27
  • just consult "one tail Z-table", for istance

    http://www.google.ru/imgres?imgurl=http://image.mathcaptain.com/cms/images/95/normal-table-large.png&imgrefurl=http://www.mathcaptain.com/probability/normal-random-variable.html&h=504&w=638&sz=17&tbnid=uWkNNcqnoZbebM:&tbnh=85&tbnw=108&zoom=1&usg=__0LIVgpES3cfUk1G_YFYEfHSb8Pw=&docid=3fz8ULfubYf6xM&sa=X&ei=pQfHUaoS6ajgBI6ogeAG&ved=0CC0Q9QEwAQ&dur=334

    p(Z = 2) = 0.9772, so probability of being defected P = 1 - p = 0.0228...

    – Dmitry Bychenko Jun 23 '13 at 14:38