Question 76064: Can someone help solve the questions that i screwed up on, on my last test. please show step by step procedure to solving these problems. ThankYou Please Help.
1. If the mean height of a women in the U.S.(20-29) is 64.0 and the standard deviation is 2.8.... What is the height that seperates the bottom 20% of women from the top 80% of women? (show standard normal graph with z-scores).
2. A random sample of 41 men were measured in height. If the mean height was 69.6 with a standard deviation of 2.426 find the 99% confidence interval for the population mean. Assume that the population standard deviation is unknown. Also assume the heights are approximately normally distributed. Round off your final answer to 2 decimal places.
3. Suppose a probability experiment consists of tossing a fair coin 200 times. Assume that heads or tails are equally likely to occur on any 1 toss. What is the probablity that the number of tails is at least 96. Round your final answer to 3 significant digits. Sketch the standard normal graph.
4. A book of 350 pages contains 525 random misprint error(s). What is the probability of exactly 2 error(s) on a randomly selected page? (Round probability to 3 decimal places. Round the meaan=((#errors)+(#pages)) to the nearest thousandth.)
b. What is the probability of no more than 3 error(s) on any one page?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. If the mean height of a women in the U.S.(20-29) is 64.0 and the standard deviation is 2.8.... What is the height that seperates the bottom 20% of women from the top 80% of women? (show standard normal graph with z-scores).
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Find the z-score that corresponds to the bottom 20% of the area under the curve:
Use InVNorm if you have a TI calculator
InvNorm(0.20)= -0.841621
Now find the raw score or "x-score" corresponding to the z-score:
-0.841621 = (x-64)/2.8
x-64= -2.356539
x = 61.64 inches
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2. A random sample of 41 men were measured in height. If the mean height was 69.6 with a standard deviation of 2.426 find the 99% confidence interval for the population mean. Assume that the population standard deviation is unknown. Also assume the heights are approximately normally distributed. Round off your final answer to 2 decimal places.
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E=z*(sigma/sqrt(n))
E=2.575829303(2.426/sqrt(41)=0.9759238861
CI: x-bar-E
CI: 69.6-0.9759238861 <=mu <= 69.6+0.9759238861
CI: (68.024,69.975923)
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The TI-83 gives the following CI:
(68.604,70.596)
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3. Suppose a probability experiment consists of tossing a fair coin 200 times. Assume that heads or tails are equally likely to occur on any 1 toss. What is the probablity that the number of tails is at least 96. Round your final answer to 3 significant digits. Sketch the standard normal graph.
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it's binomial:
n=200 ; p=1/2 ; x>=96
Use 1-binomcdf(200,0.5,95)= 0.7377
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4. A book of 350 pages contains 525 random misprint error(s). What is the probability of exactly 2 error(s) on a randomly selected page? (Round probability to 3 decimal places. Round the mean=((#errors)+(#pages)) to the nearest thousandth.)
I'll have to think about this.
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b. What is the probability of no more than 3 error(s) on any one page?
Similarly.
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Cheers,
Stan H.
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