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|What is the area under the normal curve between z = -1.0 and z = -2.0? |
|A. 0.0228 |
|B. 0.3413 |
|C. 0.1359 |
|D. 0.4772 |
| |
| |
|What is the area under the normal curve between z = 1.0 and z = 2.0? |
|A. 1.0000 |
|B. 0.7408 |
|C. 0.1359 |
|D. 0.4772 |
| |
|The mean score of a college entrance test is 500; the standard deviation is 75. The scores are normally |
|distributed. What percent of the students scored below 320? |
|A. About 50.82% |
|B. About 34.13% |
|C. About 7.86% |
|D. About 0.82% |
| |
|The mean of a normally distributed group of weekly incomes of a large group of executives is $1,000 and the |
|standard deviation is $100. What is the z-score for an income of $1,100? |
|A. 1.00 |
|B. 2.00 |
|C. 1.683 |
|D. -0.90 |
| |
|A new extended-life light bulb has an average service life of 750 hours, with a standard deviation of 50 |
|hours. If the service life of these light bulbs approximates a normal distribution, about what percent of the|
|distribution will be between 600 hours and 900 hours? |
|A. 95% |
|B. 68% |
|C. 34% |
|D. 99.7% |
| |
|The standard normal probability distribution is one which has: |
|A. A mean of 1 and any standard deviation |
|B. Any mean and a standard deviation of 1 |
|C. A mean of 0 and any standard deviation |
|D. A mean of 0 and a standard deviation of 1 |
| |
|The weekly mean income of a group of executives is $1000 and the standard deviation of this group is $100. |
|The distribution is normal. What percent of the executives have an income of $925 or less? |
|A. About 15% |
|B. About 85% |
|C. About 50% |
|D. About 23% |
| |
|What sample statistic is used to estimate a population value? |
|A. Parameter |
|B. Sampling error |
|C. Point estimate |
|D. Interval estimate |
| |
|Suppose 1,600 of 2,000 registered voters sampled said they planned to vote for the Republican candidate for |
|president. Using the 0.95 degree of confidence, what is the interval estimate for the population proportion |
|(to the nearest tenth of a percent)? |
|A. 78.2% to 81.8% |
|B. 69.2% to 86.4% |
|C. 76.5% to 83.5% |
|D. 77.7% to 82.3% |
| |
|The mean number of travel days per year for the outside salespeople employed by hardware distributors is to |
|be estimated. The 0.90 degree of confidence is to be used. The mean of a small pilot study was 150 days, with|
|a standard deviation of 14 days. If the population mean is to be estimated within two days, how many outside |
|salespeople should be sampled? |
|A. 134 |
|B. 452 |
|C. 511 |
|D. 2100 |
| |
|The proportion of junior executives leaving large manufacturing companies within three years is to be |
|estimated within 3 percent. The 0.95 degree of confidence is to be used. A study conducted several years ago |
|revealed that the percent of junior executives leaving within three years was 21. To update this study, the |
|files of how many junior executives should be studied? |
|A. 594 |
|B. 612 |
|C. 709 |
|D. 897 |
| |
|The mean weight of trucks traveling on a particular section of I-475 is not known. A state highway inspector |
|needs an estimate of the mean. He selects a random sample of 49 trucks passing the weighing station and finds|
|the mean is 15.8 tons, with a standard deviation of the sample of 3.8 tons. What is the 95 percent interval |
|for the population mean? |
|A. 14.7 and 16.9 |
|B. 13.2 and 17.6 |
|C. 10.0 and 20.0 |
|D. 16.1 and 18.1 |
| |
|Which of the following is NOT necessary to determine how large a sample to select from a population? |
|A. Level of confidence in estimating the population parameter |
|B. Size of the population |
|C. Maximum allowable error in estimating the population parameter |
|D. Estimate of the population variation |
| |
|The mean of all possible sample means is equal to the |
|A. population variance. |
|B. s2/n. |
|C. sample variance. |
|D. population mean. |
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