![]() Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.invNorm(0.70,63,5) = 65.6Ī personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Draw a new graph and label it appropriately. invNorm(area to the left, mean, standard deviation)For this problem, invNorm(0.90,63,5) = 69.4 To get this answer on the calculator, follow this step: invNorm in 2nd DISTR. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. The variable k is often called a critical value. Ninety percent of the test scores are the same or lower than k, and ten percent are the same or higher. The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Shade the area that corresponds to the 90th percentile. For each problem or part of a problem, draw a new graph. Using a computer or calculator, find P( x < 85) = 1.normalcdf(0,85,63,5) = 1 (rounds to one)The probability that one student scores less than 85 is approximately one (or 100%). Then find P( x < 85), and shade the graph. *Press 3:invNorm(*Enter the area to the left of z followed by )*Press ENTER.For this Example, the steps are 2nd Distr 3:invNorm(.6554) ENTERThe answer is 0.3999 which rounds to 0.4. The number –10 99 is way out in the left tail of the normal curve.\displaystyleArea to the left is 0.6554. In some instances, the lower number of the area might be –1E99 (= –10 99). We are calculating the area between 65 and 10 99. The number 10 99 is way out in the right tail of the normal curve. ![]() You get 1E99 (= 10ĩ9) by pressing 1, the EE key (a 2nd key) and then 99. The syntax for the instructions are as follows:normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. After pressing 2nd DISTR, press 2:normalcdf. The probability that any student selected at random scores more than 65 is 0.3446. ![]()
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