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Consider a computer test to indicate if you like Math 123 or dislike Math 123. If positive, you like the class and if negative you dislike the class. In a sample of 25 thousand people, 80% of the population indicated they liked the class. If the test is 85.5% accurate, what are the chances that the test showed the person disliked Math 123, but in reality they actually liked the course? Round your percentage to 2 decimals.

Respuesta :

Answer:

14.5%

Step-by-step explanation:

80% of the population indicated they liked the class.

Therefore: (80% X 25000)=20000 liked the class

Out of the 20,000, the test indicates that (85.5% of  20,000)=17100 are positive.

The table summarizes the computation.

[tex]\left|\begin{array}{c|c|c|c}&Positive&Negative&Total\\---&----&----&----\\Liked&17100&2900&20000\\Disliked&725&4275&5000\\\---&----&----&----\\Total&17825&7175&25000\end{array}\right|[/tex]

The table below show the various probabilities.

[tex]\left|\begin{array}{c|c|c|c}&Positive&Negative&Total\\---&----&----&----\\Liked&0.855&0.145&0.8\\Disliked&0.145&0.855&0.2\\\---&----&----&----\\Total&0.71&0.29&1\end{array}\right|[/tex]

Therefore:

P(Negative|Liked)=0.145

= 0.145X100=14.5% (to two decimal places)

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