An autoclaved aerated concrete (AAC) manufacturer packs the products into
Osvaldo Apodaca
Answered question
2021-12-11
a) No defective at all ?
b) 3 or more defectives ?
Answer & Explanation
Debbie Moore
Beginner2021-12-12Added 43 answers
Step 1
(a) Obtain the percentage of a randomly chosen box contains no defective at all.
The percentage of a randomly chosen box contains no defective at all is obtained below as follows:
Let X denotes the number of defectives in a box which follows binomial distribution with the probability of success 0.003 with the number of boxes randomly selected is 50.
That is, \(\displaystyle{n}={50},{p}={0.003},{q}={0.997}{\left(={1}-{0.003}\right)}\)
The probability distribution is given by,
\[P(X=x)=\begin{array}{c}n\\x\end{array})p^{x}(1-p)^{n-x};\ \text{here}\ x=0,1,2,...,n\ for\ 0 \le p \le 1\]
Where n is the number of trials and p is the probability of success for each trial.
The required probability is,
Use Excel to obtain the probability value for x equals 0.
Follow the instruction to obtain the P-value:
1. Open EXCEL
2. Go to Formula bar.
3. In formula bar enter the function as“=BINOMDIST”
4. Enter the number of success as 0.
5. Enter the Trails as 50.
6. Enter the probability as 0.003.
7. Enter the cumulative as False
8. Click enter.
EXCEL output:
From the Excel output, the P-value is 0.8605.
The percentage of a randomly chosen box contains no defective at all is 86.05%.
Step 2
(b) Obtain the percentage of a randomly chosen box contains 3 or more defectives.
The percentage of a randomly chosen box contains 3 or more defectives is obtained below as follows:
\(\displaystyle{P}{\left({X}\geq{3}\right)}={1}-{P}{\left({X}{ < }{3}\right)}\)
\(\displaystyle={1}-{P}{\left({X}\le{2}\right)}\)
1. Open EXCEL
2. Go to Formula bar.
3. In formula bar enter the function as“=BINOMDIST”
4. Enter the number of success as 2.
5. Enter the Trails as 50.
6. Enter the probability as 0.003.
7. Enter the cumulative as True
8. Click enter.
EXCEL output:
From the Excel output, the P-value is 0.9995.
\(\displaystyle{P}{\left({X}\geq{3}\right)}={1}-{P}{\left({X}{ < }{3}\right)}\)
\(\displaystyle={1}-{P}{\left({X}\le{2}\right)}\)
\(\displaystyle={1}-{0.9995}\)
\(\displaystyle={0.0005}\)
The percentage of a randomly chosen box contains 3 or more defectives is 0.05%.
Jim Hunt
Beginner2021-12-13Added 45 answers
Please give a more detailed answer
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