number 23 Section 3 Tho Corrnl Linse Theorem 3S3 arly of mercury (mm Hg) and the
ID: 3355987 • Letter: N
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number 23
Section 3 Tho Corrnl Linse Theorem 3S3 arly of mercury (mm Hg) and the standard deviation is 5.6 Assume the variable is bormally distributed a. If an individual is selected, find the probability that ally ind the ample the individual's pressure will be between 120 and 121.8 mm Hg b, If a sample of 30 adults is randomly selected, find the probability that the sample mean will be between 120 and 121.8 mm Hg e Why is the answer to part a so much smaller than the answer to part b? cter four thes 23. Cholesteral Content The average cholesterol content of a certain brand of eggs is 215 milligrams, and the standard deviation is 15 milligrams. Assume the vari- able is normally distributed. a. If a single egg is selected. find the probability an of a will be that the cholesterol content will be greater than 220 milligrams b. If a sample of 25 eggs is selected, find the probabil- ity that the mean of the sample will be larger than 220 milligrams Souree ing Fit ity that 24. Ages of Proofreaders At a large publishing company the mean age of proofreaders is 36.2 years, and the standard deviation is 3.7 years. Assume the variable is normally distributed. a. If a proofreader from the company is randomly se- recip e that ble is lected, find the probability that his or her age will be August inches between 36 and 37,5 years b. If a random sample of 15 proofreaders is selected, find the probability that the mean age of the proofreaders in over the the sample will be between 36 and 37.5 years. 25. TIMSS Test On the Trends in International Mathematics and Science Study (TIMSS) test in a recent year, the United States scored an average of 508 (well below South Korea, 597; Singapore, 593: Hong Kong. 572; and Japan, 570). Suppose that we take a random sample of n United States scores and that the population standard deviation is 72. If the probability s. Au- n? ths will dExplanation / Answer
We have been given the params of nomral distribution as below:
Mean = 215
Stdev = 15
We use them to standardize distribution to Z:
a. P(X>220) = P(Z> 220-215/15) = P(Z> .33) = .3707
b. P(X>220) = P(Z> 220-215/(15/sqrt(25))= P(Z> 1.65) = .0495
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