Question

Based on historical data, your manager believes that 35% of the company's orders come from first-time customers. A random sample of 161 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than than 0.19?

Answer =  (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Solution

Given that,

p = 0.35

1 - p = 1 - 0.35 = 0.65

n = 161

$\mu$_{$\hat p$} = p = 0.35

$\sigma$_{$\hat p$} =  $\sqrt$[p( 1 - p ) / n] = $\sqrt$ [(0.35 * 0.65) / 161 ] = 0.0376

P( $\hat p$ > 0.19) = 1 - P( $\hat p$ < 0.19 )

= 1 - P(( $\hat p$ - $\mu$ _{$\hat p$} ) / $\sigma$ _{$\hat p$} < (0.19 - 0.35) / 0.0376)

= 1 - P(z < -4.26)

Using z table

= 1 - 0

= 1