According to national data, about 13% of American college students earn a gradua

Question

According to national data, about 13% of American college students earn a graduate degree. Using this estimate, what is the probability that exactly 24 undergraduates in a random sample of 200 students will earn a college degree?
*Hint: use the normal approximation to the binomial distribution, where p=0.13 and q=0.87.
Please round your answer to four decimal places. According to national data, about 13% of American college students earn a graduate degree. Using this estimate, what is the probability that exactly 24 undergraduates in a random sample of 200 students will earn a college degree?
*Hint: use the normal approximation to the binomial distribution, where p=0.13 and q=0.87.
Please round your answer to four decimal places.
*Hint: use the normal approximation to the binomial distribution, where p=0.13 and q=0.87.
Please round your answer to four decimal places.

Explanation / Answer

Solution

Back-up Theory

1. If X B(n, p), for large n, {(X - np)/(npq)} can be taken as Standard Normal.

2. Since Normal Distribution is continuous, probability of exact value for X cannot be found, strictly speaking it is zero. To overcome this, we will use: P(X = 24) = P(X 24) - P(X 23) ......... (1)

3. If X N(µ, 2), P(X t) = P[Z {(t - µ)/}] where Z = {(X - µ)/} is the Standard Normal Variate whose probability can be read off from the Standard Normal Table.

Now, to solve

P(X 24) = P[Z {24 - (200 x 0.13)}/{sq.rt(200 x 0.13 x 0.87)} = P[Z (- 2/4.646)]

= P(Z - 0.43) = 0.3336 ....... (2)

Similarly,

P(X 23) = P[Z {23 - (200 x 0.13)}/{sq.rt(200 x 0.13 x 0.87)} = P[Z (- 3/4.646)]

= P(Z - 0.65) = 0.2578 ....... (3)

(1), (2) and (3) => P(X = 24) = 0.3336 - 0.2578 = 0.0758 ANSWER

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