We have learned that the Binomial distribution may be approximated by a normal d
ID: 3150559 • Letter: W
Question
We have learned that the Binomial distribution may be approximated by a normal distribution based on the Central Limit Theorem. However, the accuracy of this approximation may depend on a number of factors such as the sample size and the probability of “success”. For this problem, you are asked to evaluate the accuracy of the approximations in a few situations. Suppose that a random variable X Binomial(n, p), where X may be interpreted as the number of “success” out of n independent and identical trials. (a) Compute the following probabilities using both the Binomial distribution (which gives an exact answer) and its normal approximation (which gives an approximate answer), and compare the two answers to see if they are close or not:
(i) choose n = 10 and p = 0.2, and then compute P(X < 3) using both methods;
(ii) choose n = 10 and p = 0.4, and then compute P(X < 3) using both methods;
(iii) choose n = 50 and p = 0.2, and then compute P(X < 8) using both methods;
(iv) choose n = 50 and p = 0.4, and then compute P(X < 8) using both methods; Summarize the above results in no more than three sentences, i.e., in what cases the normal approximation may be reasonably accurate.
(b) The sample proportion ˆp can be defined as ˆp = X/n, where X is the number of success as in(a). For the cases in (i)–(iv) in (a), compute the probabilities P(ˆp < 0.3) using normal approximations. Which approximation(s) do you think may be the most accurate one based on what you have observed in (a) and why?
Explanation / Answer
a)
(i) choose n = 10 and p = 0.2, and then compute P(X < 3) using both methods;
BINOMIAL:
Note that P(fewer than x) = P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 10
p = the probability of a success = 0.2
x = our critical value of successes = 3
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 2 ) = 0.677799526
Which is also
P(fewer than 3 ) = 0.677799526 [ANSWER]
NORMAL APPROX:
We first get the z score for the critical value:
x = critical value = 2.5
u = mean = np = 2
s = standard deviation = sqrt(np(1-p)) = 1.264911064
Thus, the corresponding z score is
z = (x-u)/s = 0.395284708
Thus, the left tailed area is
P(z < 0.395284708 ) = 0.653683608 [ANSWER]
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(ii) choose n = 10 and p = 0.4, and then compute P(X < 3) using both methods;
BINOMIAL:
Note that P(fewer than x) = P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 10
p = the probability of a success = 0.4
x = our critical value of successes = 3
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 2 ) = 0.167289754
Which is also
P(fewer than 3 ) = 0.167289754 [ANSWER]
NORMAL APPROX:
We first get the z score for the critical value:
x = critical value = 2.5
u = mean = np = 4
s = standard deviation = sqrt(np(1-p)) = 1.549193338
Thus, the corresponding z score is
z = (x-u)/s = -0.968245837
Thus, the left tailed area is
P(z < -0.968245837 ) = 0.166460804 [ANSWER]
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(iii) choose n = 50 and p = 0.2, and then compute P(X < 8) using both methods;
BINOMIAL:
Note that P(fewer than x) = P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 50
p = the probability of a success = 0.2
x = our critical value of successes = 8
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 7 ) = 0.190409812
Which is also
P(fewer than 8 ) = 0.190409812 [ANSWER]
NORMAL APPROX:
We first get the z score for the critical value:
x = critical value = 7.5
u = mean = np = 10
s = standard deviation = sqrt(np(1-p)) = 2.828427125
Thus, the corresponding z score is
z = (x-u)/s = -0.883883476
Thus, the left tailed area is
P(z < -0.883883476 ) = 0.188379559 [ANSWER]
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(iv) choose n = 50 and p = 0.4, and then compute P(X < 8) using both methods;
BINOMIAL:
Note that P(fewer than x) = P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 50
p = the probability of a success = 0.4
x = our critical value of successes = 8
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 7 ) = 6.12027E-05
Which is also
P(fewer than 8 ) = 6.12027E-05 [ANSWER]
NORMAL APPROX:
We first get the z score for the critical value:
x = critical value = 7.5
u = mean = np = 20
s = standard deviation = sqrt(np(1-p)) = 3.464101615
Thus, the corresponding z score is
z = (x-u)/s = -3.608439182
Thus, the left tailed area is
P(z < -3.608439182 ) = 0.000154022 [ANSWER]
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Summarize the above results in no more than three sentences, i.e., in what cases the normal approximation may be reasonably accurate.
As we can see, the answers are closer for parts iii and iv, so it seems that the normal approximation becomes more accurate for larger sample suze. More precisely, we need np and n(1-p) to be alrge enough.
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