Question

A recent market study has determined that the probability that a young adult will be willing to try a new online financial service that your company is offering is 50%. In a random sample of 10 young adults, the probability that exactly 5 will be willing to try this new service is _________, and the approximate probability (using the normal distribution is) _________.

		0.251; 0.246
		0.246; 0.251
		0.246; 0.248
		0.249; 0.251
		0.251; 0.249

Answer 1

Binomial distribution: P(X) = nCx p^x q^n-x

n = 10

p = 0.5

q = 1 - 0.5 = 0.5

P(X = 5) = 10C5 x 0.5⁵ x 0.5⁵

= 0.246

Normal approximation: P(X < A) = P(Z < (A - mean)/standard deviation)

mean = np

= 10 x 0.5

= 5

Standard deviation = $\sqrt{npq}$

= (10x0.5x0.5)^0.5

= 1.58

P(X = 5) = P(4.5 < X < 5.5)

= P(X < 5.5) - P(X < 4.5)

= P(Z < (5.5 - 5)/1.58) - P(Z < (4.5 - 5)/1.58)

= P(Z < 0.316) - P(Z -0.316)

= 0.6240 - 0.3760

= 0.248

Ans: 0.246; 0.248