Question

Find the 10 th and the 90 th percentiles of sample proportion ^ p , if the population proportion p = 0.15 and the sample size n = 125.

Answer 1

Solution

Given that,

p = 0.15

1 - p = 1 - 0.15 = 0.85

n = 125

= p = 0.15

  [p ( 1 - p ) / n] = [(0.15 * 0.85) / 125 ] = 0.0319

a) Using standard normal table,

P(Z < z) = 10%

= P(Z < z ) = 0.10

= P(Z < -1.282 ) = 0.10  

z = -1.282

Using z-score formula,

  = z *   +  

  = -1.282 * 0.0319 + 0.15

  = 0.11

b) Using standard normal table,

P(Z < z) = 90%

= P(Z < z ) = 0.90

= P(Z < 1.282 ) = 0.90  

z = 1.282

Using z-score formula,

  = z *   +  

  = 1.282 * 0.0319 + 0.15

  = 0.19