GATE 2010-IN - 02

Question

The diameters of $10000$  ball bearings were measured. The mean diameter and standard deviation were found to be $10mm$  and    $0.05mm$  respectively. Assuming Gaussian distribution of measurements, it can be expected that the number of measurements more than $10.15mm$  will be

(A) 230  (B) 115   (C) 15   (D) 2

Answer 1

$\mu = 10$

$\sigma = 0.05$

$Z = \frac{X - \mu}{\sigma}$

$Z = \Large \frac{10.15 - 10}{0.05}$ $ = 3$
 

 

$P(Z \geq 3) =  \Large \frac{1 - P(-3 \ \leq \ Z  \ \leq 3 )}{2} $

but why dividing by 2? because the area outside -3 and 3 are equally divided to the left of -3 and to the right of 3.

 

$P(Z \geq 3) =  \Large \frac{(1 - 0.997 )}{2} $$= 0.0015 $

If $10000$ ball bearing were measure then expected number of bearings with measurements more than 10.15mm will be $10^4 \times 0.0015 = 15$.

Note : This question suggests that GATE expects student to at least remember area when standard deviation is 1,2 and 3, this is the reason $\phi(3)$ was not given in the question.

 

 

Comments

added some more details @HeadShot

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@Mk Utkarsh 

Hey , thanx. Explanation is crystal clear. I indeed analyze the explanation given by you and other mentors/ aspirants. Its helpful. Thanks Utkarsh :)

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@Mk Utkarsh

Should we remember the percentage within 1 deviation, 2 deviation and 3 deviation?

Is it necessary for GATE cse?

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Gyanu in CSE we’ve never been asked such questions afai remember, but at the same time this question is also a GATE question.