The lengths of a large stock of titanium rods follow a normal distribution with a mean ( \mu ) of 440 {mm} and a standard deviation ( \sigma ) of 1 {mm}. What is the percentage of rods whose lengths lie, between 438 {mm} and 441 {mm} ?

GATE Mechanical Engineering (ME) 2019 Shift 1 Solved Paper

Question : 26 of 65

Marks: +1, -0

The lengths of a large stock of titanium rods follow a normal distribution with a mean (

µ

) of 440

{mm}

and a standard deviation (

σ

) of

1 {mm}

. What is the percentage of rods whose lengths lie, between

438 {mm}

and

441 {mm}

Solution:

µ=440 ; σ=1

Standard normal Variable,

Z=\;{x-µ}/{σ}

For,

X=438

\ov{Z}=\;{x-µ}/{σ}=\;{438-440}/{1}=-2

Z=\;{X-µ}/{σ}=\;{441-440}/{1}=1

Now, form the normal distribution table,

P(-2)=0.4772+0.3414=81.86 \%

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