2 and 1/2 divided by 1/3
1 answer:
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The contingency tables serve to record the frequencies with which an object appears according to the measured variables, in this case two variables, each with two scales, the scratch resistance and the shock resistance, both measured as high and low.
The information is organized in such a way that the value of each box represents the total number of objects that meet the characteristics measured for each variable, for example in this case we have 70 discs with <em>high shock resistance and high scratch resistance</em>, 9 with <em>high scratch resistance and low shock resistance</em>, 16 with <em>high shock resistance and low scratch resistance</em> and finally 5 discs with <em>low scratch and shock resistance</em>.
We complete the table like this to obtain other total values:
With this information, and taking into account that the values presented are frequencies, to obtain the probability, each one must be divided into the total number of objects and multiply by 100, for this case as 100 objects, the frequency is the same percentage or probability .
Answer
a) The probability that, when selecting a random disk, its scratch resistance is high and its shock resistance is high is 70%, since this is the value that corresponds to the intersection of both variables in the table.
b) the probability that, when selecting a random disk, its resistance to scratches is high or that its resistance to impacts is high is 95% because the probability that its resistance to scratches is high is 79 % and the probability that its impact resistance is high is 86%, as one or the other requests, this is the sum of both minus the intersection, that is 79% + 86% -70% = 95%
c) the two events described are not mutually exclusive, that is, they cannot pass through, since the table informs us that of the analyzed discs 70, these two characteristics possessed, high shock resistance and high scratch resistance