Question

A recent newspaper article claimed that 18% of all sick days throughout the year are taken illegitimately. In response to this article, the Mean Corporation has randomly selected 124 employees for an annual review of the corporation's truancy rates. The officer carrying out the review has declared: "If the proportion of illegitimate sick days per year taken by you people is not less than 0.16, you will all be fired!"

Calculate the probability that the proportion of illegitimate sick days taken within the sample is less than 0.16. You may find this standard normal table useful. Give your answer as a decimal to 4 decimal places.

Answer 1

Solution

Given that,

p = 0.18

1 - p = 1 - 0.18 = 0.82

n = 124

= p = 0.18

=  [p ( 1 - p ) / n] =   [(0.18 * 0.82) / 124 ] = 0.0345

P( < 0.16)

= P[( - ) / < (0.16 - 0.18) /0.0345 ]

= P(z < -0.58)

Using z table,

= 0.2810