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. a. What is the mean and standard deviation of this distribution? 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? d.What is the probability that between 7 and 9 businesses will go bankrupt? e. What is the probability that between 7 and 15 businesses will go bankrupt?

Answer 1

A) Mean and standard deviation of this distribution :

   Mean = 10,  Std = 2.82843

B) Probability of >16 in sample will be bankrupt : 0.01444

C) Probability that exactly 14 will go bankrupt : 0.04986

D) P( x = 7, 9 ) = 0.0878

E) P( x = 7, 15 ) = 0.0757

Analysis of the Problems

n = 50

p = 20/100 = 1/5.

q = 4 / 5

Applying binomial distribution to determine the mean and std

A) Mean = E(x) = n * p

                        = 50 * 1/5 = 10

   Standard deviation = $\sqrt{np(1-p)}$  =  $\sqrt{8}$  ≈ 2.82843

B) P( x >16 )

∑$50k=17P(X=k)=\sum50k=17 \binom{50}{k}( k50​ )4^{50-k} /5^{50}$ =  0.01444.

C) Probability of ( x = 14 )

$P(X=14)=\binom{50}{14}$ *  4^36 / 5^50  ≈ 0.04986.

Applying the same formula to resolve options D and E

Hence we can conclude that the answers to your questions areas listed above:

Learn more probability : brainly.com/question/25870256

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