Question

For a particular type of fibre the number of „crimps per 10cm‟ follows a normal distribution with mean 15.1 and standard deviation 4.79. What proportion of fibre would have a „crimp per 10 cm‟ measurement of 6 or less?

Answer 1

Solution:

Let X be the number of crimps per 10cm has normal distribution with mean 15.1 and standard deviations 4.79

$\large X\sim N( \mu = 15.1, \sigma^2= 4.79^2)$

To find Proportion of fiber would have a crimp per per 10 cm measurements of 6 or less

To find P( X < 6)

$\large = P( \frac{ X - \mu}{\sigma}< \frac{ 6- \mu}{\sigma})$

= P(N(0,1) < 6 – 15.1 4.79

$\large = P( Z< \frac{ -9.1}{4.79})$

$\large = P( Z< -1.899791)$

= P( Z < -1.90)

= 0.0287. From Z table

P( X < 6) = 0.0287

0.0287 is the proportion of fiber would have crimp per 10cm measurements of 6 or less