Question

Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.

(a) Suppose n = 27 and p = 0.26. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)

np =

nq =

______ (Yes, or No), p̂ ______ (cannot, or can be) approximated by a normal random variable because ______ (np and nq do not exceed, np exceeds, np does not exceed, nq does not exceed, nq exceeds, or both np and nq exceed)

What are the values of μ_p̂ and σ_p̂? (Use 3 decimal places.)

μp̂ =

σp̂ =

(b) Suppose n = 25 and p = 0.15. Can we safely approximate p̂ by a normal distribution? Why or why not?
_____(Yes, or No), p̂  _______ (can, or cannot) be approximated by a normal random variable because _______ (np and nq do not exceed, np exceeds, np does not exceed, nq does not exceed, nq exceeds, or both np and nq exceed)
(c) Suppose n = 57 and p = 0.21. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)

np =

nq =

_____(Yes, or No), p̂  _______ (can, or cannot) be approximated by a normal random variable because _______ (np and nq do not exceed, np exceeds, np does not exceed, nq does not exceed, nq exceeds, or both np and nq exceed)

What are the values of μ_p̂ and σ_p̂? (Use 3 decimal places.)

μp̂ =

σp̂ =

Answer 1

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 μp̂ and σp̂? (Use 3 decimal places.)

μp̂ =np=27*0.26=7.02

mean=7.02

σp̂=standard deviation=sqrt(np(1-p)

=sqrt(27*0.26*(1-0.26)

σp̂= standard deviation =2.279

YES,p̂ 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, p̂ can ) be approximated by a normal random variable because  both np and nq exceed 5.

μp̂ =

n*p=

57*0.21= 11.97

mean=11.97

σp̂=standard deviation=sqrt(np(1-p)=sqrt(57*0.21*(1-0.21))

=3.075

standard deviation=3.075