A particular variety of watermelon weighs on average 23.1 pounds with a standard deviation of 1.47 pounds. Consider the sample mean weight of 95 watermelons of this variety. Assume the individual watermelon weights are independent.
Sure, you can assume the sample mean will be normally distributed (as you have quite a high n, and your population standard deviation is known) with a mean of 23.1 and a sd = your standard error = 0.151. So the probability that you'll get <23.21 from this distribution is given by the Normal cdf. For example, in Excel:
NORM.DIST(x, mean, sd, 0=pdf / 1 = cdf)
= NORM.DIST(23.21, 23.1, 0.151, 1)
Mar 8th, 2015
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