Question

According to national data, about 15% of American college students earn a graduate degree. Using this estimate, what is the probability that exactly 26 undergraduates in a random sample of 200 students will earn a college degree? Hint: Use the normal approximation to the binomial distribution, where

p = 0.15

and

q = 0.85.

(Round your answer to four decimal places.)

I think this is the formula, Im having a hard time figuring out 200C26

200C26*.15^(1-.15)^(200-26)

Answer 1

X ~ Binomial (n,p)

Where n = 0.15 , p = 200

Mean = n * p = 200 * 0.15 = 30

Standard deviation = Sqrt( n * p( 1 - p) )

= Sqrt( 200 * 0.15 * 0.85)

= 5.04975

Using normal approximation to binomial distribution and with continuity correction,

P( X = x) = P( x - 0.5 < X < x + 0.5) (Using continuity correction)

P( X = x) = P( Z = x - Mean / Standard deviation)

P( X = 26) = P( 25.5 < X < 26.5)

= P( X < 26.5) - P( X < 25.5)

= P( Z < 26.5 - 30 / 5.04975) - P( Z < 25.5 - 30 / 5.04975)

= P( Z < -0.6931) - P( Z < -0.8911)

= 0.2441 - 0.1864

= 0.0577