Peanuts (a) Since averages are less variable than individual measurements, I would expect the sample mean of 10 jars to be closer, on average, to the.

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Peanuts (a) Since averages are less variable than individual measurements, I would expect the sample mean of 10 jars to be closer, on average, to the true mean of 16.1 ounces. Thus, it is more likely that a single jar would weigh less than 16 ounces than the average of 10 jars to be less than 16 ounces.

Peanuts (b) Let X = weight of the contents of a randomly selected jar of peanuts. X is N(16.1, 0.15). P(X < 16) = normalcdf(–100, 16, 16.1, 0.15) = 0.2525. Let x(bar) = average weight of the contents of a random sample of 10 jars. is N(16.1, ). P( x(bar) < 16) = normalcdf(–100, 16, 16.1, 0.15/ 10 ) = 0.0175. This answer agrees with the answer to part (a) because this probability is much smaller than 0.2525.

Mean Texters State: What is the probability that the total number of texts in the last 24 hours is greater than 1000 for a random sample of 50 high school students? Plan: A total of 1000 texts among 50 students is the same as an average number of texts of 1000/50 = 20. We want to find P( x(bar)> 20), where x(bar) = sample mean number of texts. Since n is large (50 > 30), the distribution is approximately N(15, 35/ 50 ).

Mean Texters Do: P( x(bar)> 20) normalcdf (20, 9999, 15, 35/ 50 ) = 0.1562. Conclude: There is about a 16% chance that a random sample of 50 high school students will send more than 1000 texts in a day.