Question

Suppose the nationwide high school dropout rate for 2014 was around 10.45%. If you checked the records of 462 students in high school in 2014, what is the probability that between 8.45% and 9.98% of them ended up dropping out?

Question 2 options:

	1) 36.9864
	2) 16.1006
	3) 0.4506
	4) 0.2906
	5) 0.7094

Answer 1

Solution

Given that,

p = 0.1045

1 - p = 1 - 0.1045=0.8955

n = 462

$\mu$_{$\hat p$} = p =0.1045

$\sigma$_{$\hat p$} =  $\sqrt$[p ( 1 - p ) / n] = $\sqrt$ [(0.1045*0.8955) / 462] = 0.0142

= P( 0.0845 <$\hat p$<0.0998 )= P[(0.0845 - 0.1045) / 0.0142< ( $\hat p$ - $\mu$ _{$\hat p$} ) / $\sigma$ _{$\hat p$} < (0.0998 - 0.1045) /0.0142 ]

= P(-1.4085< z <-0.3310)

= P(z <-0.33 ) - P(z < -1.41)

Using z table,   

= 0.3701 - 0.0795

=0.2906