Question

40% of all workers of an industry are college graduates. If a sample of 200 workers is randomly selected, what is the probability that a college graduate in the sample would be

A. more than 80?

B. between 45% and 48%?

Answer 1

Solution:

$p=40\%=0.4,q=0.6,n=200 \\ X\sim Bin(n,p)$

$X\sim N(\mu,\sigma)$

$\mu=np=200(0.4)=80 \\ \sigma=\sqrt{npq}=\sqrt{200(0.4)(0.6)}=\sqrt{48}=6.93$

A. the probability that a college graduate in the sample would be more than 80

$=P(X>80) \\=P(X>80.5)$ [using continuity correction]

$=1-P(X\le 80.5) \\=1-P(Z\le \dfrac{80.5-80}{6.93}) \\=1-P(Z\le 0.072) \\=1-0.52790 \\=0.4721$

B. 45% of 200 = 90, 48% of 200 = 96

the probability that a college graduate in the sample would be between 45% and 48%

$=P(90<X<96) \\=P(X<96)-P(X<90) \\=P(X<95.5)-P(X<89.5)$ [using continuity correction]

$=P(Z<\dfrac{95.5-80}{6.93})-P(Z<\dfrac{89.5-80}{6.93}) \\=P(Z<2.24)-P(Z<1.37) \\=0.98745-0.91466 \\=0.07279$