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Let π be a continuous random variable with probability density function π of π₯ equals π over 62 π₯ for π₯ greater than or equal to 30 but less than or equal to 32 and zero otherwise.
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Find the probability that π is greater than or equal to 30.5 and less than or equal to 31.5.
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We recall first that for continuous random variable π, the probability that π₯ lies in a given interval is equal to the area under the graph of its probability density function π of π₯ between the endpoints of that interval.
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In this case, the probability density function is a straight line because π of π₯ is a linear function of π₯.
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And weβre looking for the probability that π lies between 30.5 and 31.5.
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We also recall that we can find the area under any curve, and indeed the area under a straight line, using integration.
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In general, the probability that π lies in the interval from π to π is found by evaluating the definite integral from π to π of the probability density function π of π₯ with respect to π₯.
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In this case then, we need to evaluate the integral from 30.5 to 31.5 of π over 62 π₯ with respect to π₯.
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Now, this isnβt a difficult integral to evaluate.
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But the problem is we donβt know the value of this constant π.
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Weβll need to work this out first.
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And to do so, we need to recall a probability of probability density functions, which is that for a probability density function π of π₯, the integral from negative β to positive β of π of π₯ with respect to π₯ must be equal to one, which is another way of saying that the area under the entire curve is equal to one.
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The probability density function in this question is only nonzero on the interval from 30 to 32, so we know that the integral from 30 to 32 of π over 62 π₯ with respect to π₯ must be equal to one.
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Evaluating this integral will give an equation that we can solve in order to determine the value of π.
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We recall then that to integrate a power of π₯ where the power is not equal to negative one, we increase the power, or exponent, by one and then divide by the new power.
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So the integral of π₯ with respect to π₯ is π₯ squared over two.
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And we multiply by the constant π over 62.
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This simplifies to give ππ₯ squared over 124.
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And evaluating this between the limits of 32 and 30 gives the answer one.
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Substituting in the limits of 30 and 32 gives 32 squared over 124 multiplied by π minus 30 squared over 124 multiplied by π is equal to one.
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We can factor by π and also evaluate 32 squared and 30 squared to give π multiplied by 1024 over 124 minus 900 over 124 is equal to one.
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That simplifies to π multiplied by 124 over 124 is equal to one.
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But of course, 124 over 124 is just one.
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So we have π multiplied by one equals one.
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And this tells us that the value of π is one.
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So having found the value of π, weβre now able to substitute this into the integral we wrote down earlier to enable us to find the probability that π is between 30.5 and 31.5.
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We have then the integral from 30.5 to 31.5 of one over 62 π₯ with respect to π₯.
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Integrating as before, this gives π₯ squared over 124 evaluated between 30.5 and 31.5.
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Substituting in the limits, we have 31.5 squared over 124 minus 30.5 squared also over 124.
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This evaluates to 62 over 124, which simplifies to one-half.
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So by recalling that the integral from negative β to positive β of any probability density function π of π₯ with respect to π₯ must be equal to one, we were able to determine the value of this unknown constant π and then use integration to determine the probability that π is in the given interval.
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We found the probability that π is greater than or equal to 30.5 but less than or equal to 31.5 to be one-half.