According to the Current Population Reports, published by the U.S. Bureau of the census, the mean annual alimony income received by women is $4,000. Assume a standard deviation of $7,500. Suppose 100 women are selected at random. Determine the probability that the mean annual alimony received by the 100 woman is within $500 of the population mean (sentence). Is it necessary to assume that the population of annual alimony payments is normally distributed? Explain your answer. Determine the probability that the mean annual alimony received by the 1000 woman is within $500 of the population mean (sentence). For the alimony incomes considered here, why is it necessary to take such a large sample in order to be assured of a relatively small sampling error? Judy\'s doctor is concerned that she may suffer from hypokalemia (low potassium in the blood). There is variation both in the actual potassium level and in the blood test that measures the level. Judy\'s measured potassium level varies according to the normal distribution with mu = 3.8 and sigma = 0.2. A patient is classified as hypokalemic if the potassium level is below 3.5. If a single measurement is made, what is the probability that Judy is diagnosed as hypokalemic (Sentence)?
C)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 4000 - 500 = 3500
x2 = upper bound = 4000 + 500 = 4500
u = mean = 4000
n = sample size = 1000
s = standard deviation = 7500
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -2.108185107
z2 = upper z score = (x2 - u) * sqrt(n) / s = 2.108185107
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.017507491
P(z < z2) = 0.982492509
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.964985019 [ANSWER]
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