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I have the following problem:

A manufacturer knows that on average 20% of the TVs it produces will need repair within a year. Suppose that 20 TVs are randomly selected and let $X$ be the number of TVs that will require repair.

Find then number $x$ such that the probabilit that at least $x$ of the TVs will require repairs is less than 0.5.

My answer was as follows:

$X \sim Bin(20,0.2)$ and we need to find $x$ such that $P(X\geq x)<0.5$.

Using a "binomial probability sums" table, I determine that with $n=20$ and $p=0.2$ a value of $x=3$ yields $0.4114$ and $x=4$ yields $0.6296$.

Therefore, I reasoned, that the answer is $x=3$.

The automated software we use flagged this wrong, but did not explain my mistake.

I would appreciate any help in spotting and understanding my mistake.

spaceisdarkgreen
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Confused Student
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1 Answers1

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As you said the equation is $P(X\geq x)<0.5$. Now we use the converse probability.

$$P(X\geq x)=1-P(X\leq x-1)<0.5$$

$$1-0.5 -P(X\leq x-1)<0$$

$$0.5 <P(X\leq \color{red}{x-1})\quad (*)$$

Now we look at a table for $n=20$ and $p=0.2$. And we see that $P(X<3)=0.4114$ and $P(X<\color{red}{4})=0.6296\quad (**)$. So your value for the probability is right. All it is left is to compare the red marked parts form $(*)$ and $(**)$.

$$x-1=4\Rightarrow x=5$$

callculus42
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