Consider a binomial random variable x with n = 22 and p = 0.6.
(a) Can the normal approximation be used to approximate probabilities in this case? Why or why not?
Yes, the normal approximation to the binomial probabilities can always be used.
Yes, because np and nq are greater than 5.
Yes, because np and nq are greater than 100.
No, because np and nq are less than 5.
No, because np and nq are less than 100.
(b) What are the mean μ and standard deviation σ of x? (Round your answer for standard deviation to three decimal places.)
μ =
σ =
(c) Using the correction for continuity, approximate
P(x > 8).
(Round your answer to four decimal places.)
P(x > 8) =
(a) Can the normal approximation be used to approximate probabilities in this case? Why or why not?
Yes, the normal approximation to the binomial probabilities can always be used.
Yes, because np and nq are greater than 5.
Yes, because np and nq are greater than 100.
No, because np and nq are less than 5.
No, because np and nq are less than 100.
(b) What are the mean μ and standard deviation σ of x? (Round your answer for standard deviation to three decimal places.)
μ =
σ =
(c) Using the correction for continuity, approximate
P(x > 8).
(Round your answer to four decimal places.)
P(x > 8) =
-
a) Normal approximation to Binomial distribution can be used as a rule when n > 20 or p < 0.1
or np > 5 or nq > 5
It is NOT a must to use normal approximation to Binomial distribution just because the above conditions are fulfilled.
The need for Poisson or Normal approximation to Binomial distribution arises due to the difficulty of making computations.
Since graphic/scientific calculators are now available we can use Binomial distribution and probabilities can be obtained with out any difficulty.
Unless it is specifically asked for in the question to use Poisson / normal approximation to Binomial distribution, otherwise it should NOT be used.
As such, for answering the given question, normal approximation can NOT be used. If it is desired to use CHOICE (b) can be adopted as the rule.
b) Mean = n*p = 22*0.6 = 13.2
Standard deviation = sqrt (n*p*(1-p))
= sqrt (22*0.6*0.4)
= 2.2978
c) z = (X-Mean)/SD
P(X > 8) becomes P(X > 7.5) after correction of continuity is effected.
z = (7.5 - 13.2)/2.2978 = - 2.48
The area under the standard normal curve right to the z value indicates the required probability.
P(X > 8) = 0.4934 (area between the z and mean) + 0.5000 (total area on the right side of mean)
= 0.9934
or np > 5 or nq > 5
It is NOT a must to use normal approximation to Binomial distribution just because the above conditions are fulfilled.
The need for Poisson or Normal approximation to Binomial distribution arises due to the difficulty of making computations.
Since graphic/scientific calculators are now available we can use Binomial distribution and probabilities can be obtained with out any difficulty.
Unless it is specifically asked for in the question to use Poisson / normal approximation to Binomial distribution, otherwise it should NOT be used.
As such, for answering the given question, normal approximation can NOT be used. If it is desired to use CHOICE (b) can be adopted as the rule.
b) Mean = n*p = 22*0.6 = 13.2
Standard deviation = sqrt (n*p*(1-p))
= sqrt (22*0.6*0.4)
= 2.2978
c) z = (X-Mean)/SD
P(X > 8) becomes P(X > 7.5) after correction of continuity is effected.
z = (7.5 - 13.2)/2.2978 = - 2.48
The area under the standard normal curve right to the z value indicates the required probability.
P(X > 8) = 0.4934 (area between the z and mean) + 0.5000 (total area on the right side of mean)
= 0.9934