The testing division of a chip manufacturing company tests all manufac
ukelangf0
Answered question
2021-12-02
Answer & Explanation
Befory
Beginner2021-12-03Added 19 answers
Step 1
Introduction to Geometric distribution:
Suppose there are independent trials and the probability of success is same in each trial, then the probability of x trials for preceding the first success is \(\displaystyle{q}^{{{x}-{1}}}{p}\).
The outcome of each trial can be classified as either a “success” or a “failure”. The numerical value “1” is assigned to each success and “0” is assigned to each failure.
Moreover, the probability of getting a success in each trial, p, remains a constant for all the n trials. Denote the probability of failure as q. As success and failure are mutually exclusive, \(\displaystyle{q}={1}-{p}\).
Let the random variable X denote the number of trials. Thus, X can take any of the values 0,1,2,…,
Then, the probability distribution of X is a Geometric distribution with parameters (p) and the probability mass function (pmf) of X is given as:
\(f(x)=\begin{cases}q^{x-1}p, & x = 0,1,... and\ r>0;0<p<1; q=1-p\\0, & otherwise\end{cases}\)
Step 2
Calculation:
In the given random experiment, the first defective chip is tested on a given day appears after the 10th tested chip with probability 0.9. Since, the first defective chip appears after the 10th chip, the probability of 0.9 is considered as the probability of success by assuming 10th chip as the starting chip.
Consider the event that the defective chip appears as a “success”. It is known that the the probability of success in each trial is \(\displaystyle{p}={0.9}\). Then \(\displaystyle{q}={1}-{0.9}={0.1}\).
Consider X as the number of trials preceding the first success. Then, X is a Geometric distribution with parameters \(\displaystyle{\left({p}={0.9}\right)}\).
The probability that the first defective chip appears between the 20th and the 30th chips excluding the 20th and the 30th chips on a given day is obtained as 0.000 from the calculation given below:
\(\displaystyle{f{{\left({20}{<}{X}{<}{30}\right)}}}={P}{\left({21}\le{X}\le{29}\right)}\)
\(\displaystyle={P}{\left({12}\le{X}\le{20}\right)}\)
\(\displaystyle={\sum_{{{x}={12}}}^{{{20}}}}{\left({0.1}\right)}^{{{x}-{1}}}{\left({0.9}\right)}\)
\(\displaystyle={\left[{\left({\left({0.1}\right)}^{{{11}}}{\left({0.9}\right)}\right)}+{\left({\left({0.1}\right)}^{{{12}}}{\left({0.9}\right)}\right)}+{\left({\left({0.1}\right)}^{{{13}}}{\left({0.9}\right)}\right)}+{\left({\left({0.1}\right)}^{{{14}}}{\left({0.9}\right)}\right)}+{\left({\left({0.1}\right)}^{{{15}}}{\left({0.9}\right)}\right)}+{\left({\left({0.1}\right)}^{{{16}}}{\left({0.9}\right)}\right)}+{\left({\left({0.1}\right)}^{{{17}}}{\left({0.9}\right)}\right)}+{\left({\left({0.1}\right)}^{{{18}}}{\left({0.9}\right)}\right)}+{\left({\left({0.1}\right)}^{{{19}}}{\left({0.9}\right)}\right)}\right]}\)
\(\displaystyle={0.000}\)
Thus, the probability that the first defective chip appears between the 20th and the 30th chips excluding the 20th and the 30th chips on a given day is 0.000.
New Questions in High school statistics
Read carefully and choose only one option
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