Answer to Question #127740 in Statistics and Probability for Alicia Casillas

Question

It is very difficult for small businesses to be successful. The Small Business Administration estimates that 20 percent will dissolve or go bankrupt within two years. A sample of 50 new businesses is selected. What is the mean and standard deviation of this distribution? a. b. What is the probability that more than 16 in the sample will go bankrupt? c. What is the probability that exactly 14 will go bankrupt?

Answer 1

Solution:

n=50

20%/100%=1/5.

p=1/5.

This is binomial distribution with parameters n=50, p=1/5.

Then P(X=k)=$\binom{n}{k}$ p^k(1-p)^n-k

P(X=k)=$\binom{50}{k}$ (1/5)^k(4/5)^50-k=$\binom{50}{k}$4^50-k/5⁵⁰.

Here $\binom{n}{k}$ =n!/(k!(n-k)!) is binomial coefficient

a. Mean: E[X]=np=50$\cdot$1/5=10.

Standart deviation: $\sigma=\sqrt{np(1-p)}=\sqrt{50\cdot 1/5\cdot 4/5}=\sqrt{8}=2\sqrt{2}\approx2.82843$

b. P(X>16)=$\sum$ ⁵⁰_k=17P(X=k)=$\sum$⁵⁰_k=17 $\binom{50}{k}$4^50-k/5⁵⁰$\approx0.01444$.

The probability that more than 16 in the sample will go bankrupt is approximately equal 0.01444

c. P(X=14)=$\binom{50}{14}$ $\cdot$ 4^36/5^50$\approx0.04986$.

The probability that exactly 14 will go bankrupt is approximately equal =0.04986.

We used for calculations Maple and Wolfram Alpha.