Question

Depths of pits on a corroded steel surface are normally distributed with mean 820 μm and standard deviation 29 μm.

a) Find the 10th percentile of pit depths.

b) A certain pit is 780 μm deep. What percentile is it on? (Round up the final answer to the nearest whole number.)

Answer 1

Solution:-

Given that,

mean = = 820 um

standard deviation = = 29 um

a) Using standard normal table,

P(Z < z) = 10%

= P(Z < z ) = 0.10

= P(Z < -1.28 ) = 0.10  

z = -1.28

Using z-score formula,

x = z * +

x = -1.28 * 29 + 820

x = 782.88

x = 783 um

b) P(x < 780)

= P[(x - ) / < (780 - 820) / 29]

= P(z < -1.38)

Using z table,

= 0.0838

The percentage is = 8.38%

A percentile = 8 th percentile