Answer to Question #206523 in Statistics and Probability for Vân

Question

A successful venture-capital firm notes that it provides financing for only 30% of the proposals it reviews. This year it reviews 120 proposals.

a/ What is the probability that at least 30 of the 120 proposals submitted will receive financing?

b/ What is the probability that the number of proposals receiving financing will be between 40 and 60?

Answer 1

Probability of providing financing , p = 0.30

Number of proposal reviewed n = 120

Therefore, q = 1-p = 0.70

Mean $\,u=np=120\times 0.30=36$

Standard Deviation $\sigma = \sqrt{npq}=\sqrt{120\times 0.30\times 0.70}=5.02$

(a) Probability of atleast 30 of 120 proposals submitted receive financing $P(X\geq30)$

We want to make continuity correction

$P(X\geq 30 )=P(X\geq 29.5)$

The calculation reduced to normal probability calculation

$P(X\geq 29.5) = P(\frac{X-\mu}{\sigma }\geq \frac{29.5-36}{5.02})$

$=P(X\geq -1.295)\\=0.9023\ [\text{From standard normal table}]$

(b) Probability of number of proposals receive financing between 40 and 60

$P(40<X<60)$

By continuity correction

$P(40<X<60)=P(40.5<X<59.5)$

$P(40.5<X<59.5)=P(\frac{40.5-36}{5.02}<\frac{X-\mu}{\sigma}<\frac{59.5-36}{5.02})$

$=P(0.9<Z<4.68)\\=P(Z<4.68)-P(Z<0.9)\\=1-0.815\\=0.185$