Answer to Question #170181 in Statistics and Probability for andrew

Question #170181

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

(a) Suppose n = 28 and p = 0.18. Can we approximate by a normal distribution? Why? (Use 2 decimal places.)

np = nq =

, be approximated by a normal random variable because .


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

(b) Suppose n = 25 and p = 0.15. Can we safely approximate by a normal distribution? Why or why not?

, be approximated by a normal random variable because .


(c) Suppose n = 40 and p = 0.39. Can we approximate by a normal distribution? Why? (Use 2 decimal places.)

np = nq =

, be approximated by a normal random variable because .


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


1
Expert's answer
2021-03-09T08:08:21-0500

If "X\\sim Bin(n, p)"and if "n" is large and/or "p" is close to "1\/2," then "X" is approximately "N(np, npq)," where "q=1-p."

In practice, the approximation is adequate provided that both "np\\geq10" and "nq\\geq 10," since there is then enough symmetry in the underlying binomial

distribution.

Sometimes consider the condition "np>5" and "nq>5."

If the binomial probability histogram is not too skewed, "X" has approximately a normal distribution with "\\mu=np" and "\\sigma=\\sqrt{npq}"

(a) "n=28, p=0.18"

"q=1-p=1-0.18=0.82"

Check the condition "np>5" and "nq>5"

"np=28(0.18)=5.04>5"

"nq=28(0.82)=22.96>5"

We can safely approximate "\\hat{p}"  by a normal distribution


"\\mu=np=28(0.18)=5.04"

"\\sigma=\\sqrt{npq}=\\sqrt{28(0.18)(0.82)}=2.033"



Check the condition "np\\geq10" and "nq\\geq10"

"np=28(0.18)=5.04<10"

We cannot safely approximate "\\hat{p}"  by a normal distribution.


(b) "n=25, p=0.15"

"q=1-p=1-0.15=0.85"

Check the condition "np>5" and "nq>5"

"np=25(0.15)=3.75<5"

We cannot safely approximate "\\hat{p}"  by a normal distribution.


Check the condition "np\\geq10" and "nq\\geq10"

"np=25(0.15)=3.75<10"

We cannot safely approximate "\\hat{p}"  by a normal distribution.


(c) "n=40, p=0.39"

"q=1-p=1-0.39=0.61"

Check the condition "np>5" and "nq>5"

"np=40(0.39)=15.6>5"

"nq=40(0.61)=24.4>5"

We can safely approximate "\\hat{p}"  by a normal distribution


"\\mu=np=40(0.39)=15.6"

"\\sigma=\\sqrt{npq}=\\sqrt{40(0.39)(0.61)}=3.085"



Check the condition "np\\geq10" and "nq\\geq10"

"np=40(0.39)=15.6\\geq10"

"nq=40(0.61)=24.4\\geq10"

We can safely approximate "\\hat{p}" by a normal distribution


"\\mu=np=40(0.39)=15.6"

"\\sigma=\\sqrt{npq}=\\sqrt{40(0.39)(0.61)}=3.085"




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