Suppose we have a binomial experiment in which success is defined to be a particular quality...

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Suppose we have a binomial experiment in which success isdefined to be a particular quality or attribute that interestsus.
(a) Suppose n = 27 and p = 0.26. Can weapproximate p? by a normal distribution? Why? (Use 2decimal places.)
np =
nq =
(Yes, or No), p? (cannot, or can be)approximated by a normal random variable because __ (np and nqdo 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 ?_p? and?_p?? (Use 3 decimal places.)
?_p? =
?_p? =

(b) Suppose n = 25 and p = 0.15. Can we safelyapproximate p? by a normal distribution? Why or whynot?
_(Yes, or No), p? _ (can, or cannot)be approximated by a normal random variable because ___ (np andnq do not exceed, np exceeds, np does not exceed, nq does notexceed, nq exceeds, or both np and nq exceed)
(c) Suppose n = 57 and p = 0.21. Can weapproximate p? by a normal distribution? Why? (Use 2decimal places.)
np =
nq =

_(Yes, or No), p? _ (can, or cannot)be approximated by a normal random variable because ___ (np andnq do not exceed, np exceeds, np does not exceed, nq does notexceed, nq exceeds, or both np and nq exceed)

What are the values of ?_p? and?_p?? (Use 3 decimal places.)
?_p? =
?_p? =

Answer & Explanation Solved by verified expert

4.5 Ratings (881 Votes)

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.

?p? =

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

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