A recent study by the EPA has determined that the amount of contaminants in Minnesota lakes (in parts per million) is normally distributed with mean 64 ppm and variance 17.6. Suppose 35 lakes are randomly selected and sampled. What is the probability that the sample average amount of contaminants is (a) Above 72 ppm? (b) Between 64 and 72 ppm? (c) Exactly 64 ppm.

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Himin1945 · Accepted Answer

Step 1 Let us define the random variable X follows normal distribution mean 64 ppm and variance 17.6. That is, μ=64 and σ=17.6=4.1952 a) The probability that the sample average amount of contaminants above 72 ppm is, P(x―&amp;gt;72)=P(x―−μσn&amp;gt;72−μσn) =P(z&amp;gt;72−644.195235) =P(z&amp;gt;11.2816) =1−P(z≤11.2816) =0from the excel function,=1−NORM .DIST(11.2816,0,1,TRUE) Step 2 b) The probability that the sample average amount of contaminants between 64 and 72 ppm is, P(64&amp;lt;x―&amp;lt;72)=P(64−μσn&amp;lt;x―−μσn&amp;lt;72−μσn) =P(64−644.195235&amp;lt;z&amp;lt;72−644.195235) =P(0&amp;lt;z&amp;lt;11.2816) =P(z&amp;lt;11.2816)−P(z&amp;lt;0) =0.5from the excel function,=NORM .DIST (11.2816,0,1,TRUE)-NORM.DIST(0,0,1,TRUE) c) Note: For continuous distributions, the probability at particular point is zero. That is, P(X=a)=0 Here, the given distribution is normal distribution. Hence, the probability that the sample average amount of contaminants exactly 64 ppm is 0.

A recent study by the EPA has determined that the amount of contaminan

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