SolutionA:
np=27*0.26=7.02
nq=27*(1-0.26)=19.98
np,nq>=5
normal distribution can be used as an approxiamtion to
binomial.
What are the values of ?p? and ?p?? (Use 3 decimal places.)
?p? =np=27*0.26=7.02
mean=7.02
?p?=standard deviation=sqrt(np(1-p)
=sqrt(27*0.26*(1-0.26)
?p?= standard deviation =2.279
YES,p? can be pproximated by a normal random variable because
both np and nq exceed 5
Solutionb:
n=25
p=0.15
q=1-p=1-0.15=0.85
np=25*0.15=3.75
nq=25*(0.85)=21.25
NO,p^ cannot) be approximated by a normal random
variable because np does not exceed 5
Solutionc:
n=57
p=0.21
q=1-p=1-0.21=0.79
np=57*0.21= 11.97
nq=57*0.79=45.03
np,np >=5
use normal approximation
YES, p? can ) be approximated by a normal random variable
because both np and nq exceed 5.
n*p=
57*0.21= 11.97
mean=11.97
?p?=standard deviation=sqrt(np(1-p)=sqrt(57*0.21*(1-0.21))
=3.075
standard deviation=3.075
