Q:

A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the number of underfilled will be: Note: This question uses a tolerance of .0001 for grading. Make sure your answers are entered to the fifth decimal place.(Example: .00123) (a) exactly 1 Pr= (b) exactly 3 Pr= (c) exactly 9 Pr= (d) at least 5 Pr=

Accepted Solution

A:
Answer:a) [tex]P(X = 1) = 0.38742[/tex]b) [tex]P(X = 3) = 0.05740[/tex]c) [tex]P(X = 9) = 0.00000[/tex]d) [tex]P(X \geq 5) = 0.00163[/tex]Step-by-step explanation:For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.Binomial probability distribution:The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]And p is the probability of X happening.In this problemThere are 10 containers, so [tex]n = 10[/tex].A food-packaging apparatus underfills 10% of the containers, so [tex]p = 0.1[/tex].a) This is P(X = 1)[tex]P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742[/tex]b) This is P(X = 3)[tex]P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740[/tex]c) This is P(X = 9)[tex]P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000[/tex]d) This is [tex]P(X \geq 5)[/tex].Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:[tex]P(X < 5) + P(X \geq 5) = 1[/tex][tex]P(X \geq 5) = 1 - P(X < 5)[/tex]In which[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex][tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex][tex]P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868[/tex][tex]P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742[/tex][tex]P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937[/tex][tex]P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740[/tex][tex]P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742[/tex]So[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837[/tex]Finally[tex]P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163[/tex]