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 p̂ by a normal distribution? Why? (Use 2 decimal places.)
np = nq =
, p̂ be approximated by a normal random variable because .
What are the values of 𝜇p̂ and 𝜎p̂? (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?
, p̂ be approximated by a normal random variable because .
(c) Suppose n = 40 and p = 0.39. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)
np = nq =
, p̂ be approximated by a normal random variable because .
What are the values of 𝜇p̂ and 𝜎p̂? (Use 3 decimal places.) 𝜇p̂ = 𝜎p̂ =
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
"\\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
"\\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
"\\sigma=\\sqrt{npq}=\\sqrt{40(0.39)(0.61)}=3.085"
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