Respuesta :
Answer:
The probability is [tex]P(X \le 9) =0.9138 [/tex]
Step-by-step explanation:
From the question we are told that
The probability of the television passing the test is p = 0.95
The sample size is n = 10
Generally the comprehensive testing process for all essential functions follows a binomial distribution
i.e
[tex]X \~ \ \ \ B(n , p)[/tex]
and the probability distribution function for binomial distribution is
[tex]P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}[/tex]
Here C stands for combination hence we are going to be making use of the combination function in our calculators
Generally the probability that at least 9 pass the test is mathematically represented as
[tex]P(X \le 9) = P(X = 9) + P(X = 10 )[/tex]
=> [tex]P(X \le 9) = [^{10}C_9 * (0.95)^9 * (1- 0.95)^{10-9}]+ [^{10}C_{10} * {0.95}^{10} * (1- 0.95)^{10-10}][/tex]
=> [tex]P(X \le 9) = [10 * 0.6302 * 0.05 ]+ [1 *0.5987 * 1 ] [/tex]
=> [tex]P(X \le 9) =0.9138 [/tex]
Using the binomial distribution, it is found that there is a 0.9138 = 91.38% probability that at least 9 pass the test.
For each television, there are only two possible outcomes. Either they pass the tests, or they do not. The probability of a television passing the test is independent of any other television, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- There are 10 televisions, thus [tex]n = 10[/tex].
- Each has a 0.95 probability of passing the tests, thus [tex]p = 0.95[/tex].
The probability that at least 9 pass the tests is:
[tex]P(X \geq 9) = P(X = 9) + P(X = 10)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 9) = C_{10,9}.(0.95)^{9}.(0.05)^{1} = 0.3151[/tex]
[tex]P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987[/tex]
Then
[tex]P(X \geq 9) = P(X = 9) + P(X = 10) = 0.3151 + 0.5987 = 0.9138[/tex]
0.9138 = 91.38% probability that at least 9 pass the test.
A similar problem is given at https://brainly.com/question/24863377
