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Problem 85: A manufacturer of external hard drives claims that only 10% of his drives require repairs within the warranty peroid of 12 months.

Problem 85: A manufacturer of external hard drives claims that only 10% of his drives require repairs within the warranty peroid of 12 months.


a. If 5 of 20 of his drives required repairs within the first year, does this tend to support or refuse the claim? b. If 3 of 20 of his drives required repairs within the first year, does this tend to support or refuse the claim?

Solution:


a. If 5 of 20 of his drives required repairs within the first year
Given that $n=20,$ and $p=0.3$
$\sum _{x=5}^{20}b(x;n,p)=\sum _{x=5}^{20}b(x;20,0.1)=1-\sum _{x=0}^{4}b(x;20,0.1)$
$ = 1 - [b(0;20,0.1)+b(1;20,0.1)+b(2;20,0.1)+b(3;20,0.1)+b(4;20,0.1)]$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(0;20,0.1) = \left(\begin{array}{c}20\\ 0\end{array}\right){0.1}^{0}(1-0.1)^{20-0} = 0.121577$
$b(1;20,0.1) = \left(\begin{array}{c}20\\ 1\end{array}\right){0.1}^{1}(1-0.1)^{20-1} = 0.27017$
$b(2;20,0.1) = \left(\begin{array}{c}20\\ 2\end{array}\right){0.1}^{2}(1-0.1)^{20-2} = 0.28518$
$b(3;20,0.1) = \left(\begin{array}{c}20\\ 3\end{array}\right){0.1}^{3}(1-0.1)^{20-3} = 0.19012$
$b(4;20,0.1) = \left(\begin{array}{c}20\\ 4\end{array}\right){0.1}^{4}(1-0.1)^{20-4} = 0.08978$
$\sum _{x=5}^{20}b(x;n,p) = 1 - [b(0;20,0.1)+b(1;20,0.1)+b(2;20,0.1)+b(3;20,0.1)+b(4;20,0.1)]$
$= 1-(0.121577+0.27017+0.28518+0.19012+0.08978) = 1-0.9568 =0.0432$
since this probability is very small, it would seem reasonable to refuse the hard drive manufacturer's claim.
b . If 3 of 20 of his drives required repairs within the first year
Given that $n=20,$ and $p=0.3$
$\sum _{x=3}^{20}b(x;n,p)=\sum _{x=3}^{20}b(x;20,0.1)=1-\sum _{x=0}^{2}b(x;20,0.1)$
$ = 1 - [b(0;20,0.1)+b(1;20,0.1)+b(2;20,0.1)]$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(0;20,0.1) = \left(\begin{array}{c}20\\ 0\end{array}\right){0.1}^{0}(1-0.1)^{20-0} = 0.121577$
$b(1;20,0.1) = \left(\begin{array}{c}20\\ 1\end{array}\right){0.1}^{1}(1-0.1)^{20-1} = 0.27017$
$b(2;20,0.1) = \left(\begin{array}{c}20\\ 2\end{array}\right){0.1}^{2}(1-0.1)^{20-2} = 0.28518$
$\sum _{x=5}^{20}b(x;n,p) = 1 - [b(0;20,0.1)+b(1;20,0.1)+b(2;20,0.1)]$
$ = 1-(0.121577+0.27017+0.28518) = 1-0.6769 =0.3231$
since this probability is quite large, it would seem reasonable to support the hard drive manufacturer's claim.

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