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  3. suppose we have a binomial experiment in which suc

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

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Basic Math

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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