Question

In a recent poll, 45% of survey respondents said that, if they only had one child, they would prefer the child to be a boy. Suppose you conducted a survey of 160 randomly selected students on your campus and find that 82 of them would prefer a boy. Complete parts (a) and (b) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) Use the normal approximation to the binomial to approximate the probability that, in a random sample of 160 students, at least 82 would prefer a boy, assuming the true percentage is 45%. . The probability that at least 82 students would prefer a boy is (Round to four decimal places as needed.)

Answer 1

Solution :

Given that,

p = 0.45

q = 1 - p =1 -0.45 =0.55

n = 160

Using binomial distribution,

$\mu$ = n * p = 160 *0.45=72

$\sigma$ = $\sqrt$ n * p * q = $\sqrt$ 160*0.45*0.55=6.2929

Using continuity correction

,P(x >82) = 1 - P(x <81.5 )

= 1 - P((x - $\mu$ ) / $\sigma$ < (81.5 -72) / 6.2929)

= 1 - P(z <1.51 )   

Using z table   

=1-0.9345

=0.0655

probability= 0.0655