Question

15. A report indicated that 43% of adults had received a "phishing" contact (a bogus email that replicates an authentic site for the purpose of stealing personal information such as account numbers and passwords). Suppose a random sample of 800 adults is obtained. a) What is the probability that no more than 40% have received a phishing contact? b) What is the probability that between 37% and 42% have received a phishing contact?

Answer 1

Solution

Given that,

p = 0.43

1 - p = 1 - 0.43 = 0.57

n = 800

$\mu$_{$\hat p$} = p = 0.43

$\sigma$_{$\hat p$} =  $\sqrt$[p ( 1 - p ) / n] =  $\sqrt$ [(0.43 * 0.57) / 800 ] = 0.0175

a) P( $\hat p$ < 0.40)

= P[( $\hat p$ - $\mu$ _{$\hat p$} ) / $\sigma$ _{$\hat p$} < (0.40 - 0.43) / 0.0175 ]

= P(z < -1.71)

Using z table,

= 0.0436

b) P( 0.37 < $\hat p$ < 0.42 )

= P[(0.37 - 0.43) / 0.0175 < ( $\hat p$ - $\mu$ _{$\hat p$} ) / $\sigma$ _{$\hat p$} < (0.42 - 0.43) / 0.0175 ]

= P(-3.43 < z < -0.57)

= P(z < -0.57) - P(z < -3.43)

Using z table,   

= 0.2843 - 0.0003

= 0.2840