It is estimated that 10% of the vehicles entering Canada from the United States carry undeclared goods. Assuming the data can be modelled using a normal approximation, determine the probability that a search of 500 randomly selected vehicles will find fewer than 50 with undeclared goods
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It is estimated that 10% of the vehicles entering Canada from the United States carry undeclared goods. Assuming the data can be modelled using a normal approximation, determine the probability that a search of 500 randomly selected vehicles will find fewer than 50 with undeclared goods.
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- An average of many different studies of handedness indicate that in a random sample of adults,14percent of men are left-handed and 6 percent of women are left-handed. (Assume the rest are right-handed.) Suppose you have a sample of 250 women and 100 men, in which the numbers of left-handed and right-handed people reflect the average percentages. Make a two-way table that shows the distribution you would find.Suppose that an ice cream dispenser fills up an average of 200 mL per cup. If it is in normal distribution and the SD=15mL, 1. Find the fraction of cups that will be filled with more than 224 mL. 2. Find the probability that the cup will be filled between 191mL & 209mL. 3. Below what mL will have the smallest 25% of the ice cream?Toward the middle of the harvesting season, peaches for canning come in three types, early, late, and extra late, depending on the expected date of ripening. During a certain week, the data to the right were recorded at a fruit delivery station. Complete parts (a) through (d) below. 18 trucks (Type a whole number.) (b) How many carried only extra late? 17 trucks (Type a whole number.) (c) How many carried only one type of peach? 28 trucks went out carrying early peaches; 60 carried late peaches; (a) How many trucks carried only late variety peaches? trucks (Type a whole number.) 46 carried extra late peaches; 19 carried early and late; 25 carried late and extra late; 6 carried early and extra late; 2 carried all three; 5 carried only figs (no peaches at all).
- According to a study conducted by a statistical organization, the proportion of people who are satisfied with the way things are going in their lives is 0.78. Suppose that a random sample of 100 people is obtained. Complete parts (a) through (e) below. different people feel differently regarding their satisfaction. D. The sample proportion p is a random variable because the value of p represents a random person included in the sample. The variability is due to the fact that people may not be responding to the question truthfully. (c) Describe the sampling distribution of p, the proportion of people who are satisfied with the way things are going in their life. Be sure to verify the model requirements. Since the sample size is no more than 5% of the population size and np(1 - p) = 17.16 2 10, the distribution of p is approximately normal with Ha = 0.780 and On = 0.041 . (Round to three decimal places as needed.) (d) In the sample obtained in part (a), what is the probability that the…The price of a share of stock divided by the company's estimated future earnings per share is called the P/E ratio. High P/E ratios usually indicate "growth" stocks, or maybe stocks that are simply overpriced. Low P/E ratios indicate "value" stocks or bargain stocks. A random sample of 51 of the largest companies in the United States gave the following P/E ratios†. 11 35 19 13 15 21 40 18 60 72 9 20 29 53 16 26 21 14 21 27 10 12 47 14 33 14 18 17 20 19 13 25 23 27 5 16 8 49 44 20 27 8 19 12 31 67 51 26 19 18 32 (a) Use a calculator with mean and sample standard deviation keys to find the sample mean x and sample standard deviation s. (Round your answers to four decimal places.) x = s = (b) Find a 90% confidence interval for the P/E population mean ? of all large U.S. companies. (Round your answers to one decimal place.) lower limit upper limit (c) Find a 99% confidence interval for the P/E population mean ? of all large U.S.…Recall that Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.306. Now suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 219 numerical entries from the file and r = 49 of the entries had a first nonzero digit of 1. Let p represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1.Test the claim that p is less than 0.306. Use ? = 0.05. (a) What is the level of significance?State the null hypothesis H0 and the alternate hypothesis H1 . H0 : p H1 : p (b) What sampling distribution will you use? The Student's tThe standard normal since np…
- A researcher believes that 52.5% of people who grew up as the only child have an IQ score over 100. However, unknown to the researcher, this figure is actually 50%, which is the same as in the general population.To attempt to find evidence for the claim, the researcher is going to take a random sample of 400 people who grew up as the only child. Let p be the proportion of people in the sample with an IQ score above 100. Answer the following: a) Find the mean of p: b)Find the standard deviation of p: c) Compute an approximation for P(p ≥ 0.525), which is the probability that there will be 52.5% or more people with IQ scores over 100 in the sample. Round your answer to four decimal places.About 3% of the population has a particular genetic mutation. 700 people are randomly selected.Find the mean for the number of people with the genetic mutation in such groups of 700. (Round to 2 decimal places if possible.)Sleep apnea is a condition in which the sufferers stop breathing momentarily while they are asleep. This condition results in lack of sleep and extreme fatigue during waking hours. A current estimate is that 9.7 million out of the 312.7million Americans suffer from sleep apnea, or approximately 3.1%A safety commission is concerned about the percentage of commercial truck drivers who suffer from sleep apnea. They do not have any reason to believe that it would be higher or lower than the population’s percentage. To test the claim that the percentage of commercial truck drivers who suffer from sleep apnea is not 3.1%, a simple random sample of 406commercial truck drivers is examined by a medical expert, who concludes that 55 suffer from sleep apnea. Does this evidence support the claim that the percentage of commercial truck drivers who suffer from sleep apnea is not 3.1%?Use a 0.05 level of significance. Step 3 of 3 : Draw a conclusion and interpret the decision.
- According to a study conducted by a statistical organization, the proportion of people who are satisfied with the way things are going in their lives is 0.78.Suppose that a random sample of 100 people is obtained. Complete parts (a) through (e) below.In your asnwer, please explain how to solve using the help of excel if possible.About 7% of the population has a particular genetic mutation. 100 people are randomly selected. Find the mean for the number of people with the genetic mutation in such groups of 100. (Round to 2 decimal places if possible.)
