Question

Suppose that you and two friends go to a​ restaurant, which last month filled approximately 86% of the orders correctly.

(a) What is the probability that all three orders will be filled​ correctly?

The probability is 0.6361

(b) What is the probability that none of the three orders will be filled​ correctly?

The probability is 0.0022

(c) What is the probability that at least two of the three orders will be filled​ correctly?

The probability is 0.9466

(d) What are the mean and standard deviation of the binomial distribution used in​ (a) through​ (c)? Interpret these values.

The mean is= ?

(Round to four decimal places)

Answer 1

Given the

probability that of order will correctly is p=0.86

Let X be the number of correct orders.

(a)The probability that all three orders will be filled​ correct:

         

Hence the probability that all the three orders will fill correctly is 0.6361

(b) The probability that none of the three orders will be filled​ correctly:

That is

P( X = 0) = ³C₀ 0.86⁰ 0.14³

Hence theprobability that none of the three orders will be filled​ correctly is 0.0022

(c) The probability that at least two of the three orders will be filled​ correctly:

That is

P( X >= 2) = P( x = 2) + P( x = 3)

= ³C₂ 0.86² 0.14 + ³C₃ 0.86³ 0.14⁰

= 0.9466

Hence the probability that at least two of the three orders will be filled​ correctly is 0.9466

(d) mean and standard deviation of the binomial distribution :

Mean = n * p

= 3 * 0.86

= 2.58

Standard deviation = Sqrt( n p ( 1-p) )

= Sqrt( 3 * 0.86 * 0.14 )

= 0.60099