1) 85% of students are in favor of freezing CUNY tuition. In a random sample of...
1. (5 points) Suppose Z is a random variable that follows the standard normal distribution. a) Find P(Z > 0.45). b) Find P(0.7 SZ 1.6). c) Find 20.09. d) Find the Z-score for having area 0.18 to its left under the standard normal curve. e) Find the value of z such that P(-2SZS2) -0.5. 3. (4 points) The scores on a test are normally distributed with a mean of 75 and a standard deviation of 8. a) Find the proportion...
1. (5 points) Suppose Z is a random variable that follows the standard normal distribution. a) Find P(Z > 0.45). b) Find P(0.7 SZ 1.6). c) Find 20.09. d) Find the Z-score for having area 0.18 to its left under the standard normal curve. e) Find the value of z such that P(-2SZS2) -0.5.
40 B 1. (5 points) Suppose Z is a random variable that follows the standard normal distribution. a) Find P(Z > 0.45). b) Find P(0.7 SZ 1.6). c) Find 20.09. d) Find the Z-score for having area 0.18 to its left under the standard normal curve. e) Find the value of z such that P(-2SZS2) -0.5.
A random sample of 85 students finds that they, on average, they get 6.5 hours of sleep a night, with a sample standard deviation of 7 hours. Would you use the z-distribution or the t-distribution to construct a confidence interval? How do you know? Construct a 95% confidence interval for the population mean. How do you interpret this interval? Is it likely that students actually get 7 hours of sleep a night? How can you tell?
1. We believe that 85% of the population of all Business Statistics students consider statistics to be an exciting subject. Suppose we randomly and independently selected 40 students from the population. If the true percentage is really 85%, find the probability of observing 39 or more students who consider statistics to be an exciting subject. Round to six decimal places. 0.001502 0.012107 0.987893 0.010604 2. A physical fitness association is including the mile run in its secondary-school fitness test. The...
A random sample of 100 students is taken from the
population of all part-time students in the U.S., for which the
overall population of females is 0.60.6. Start by determining what
is given in the original statement.
Probabilities: Proportion Sampling Distributions If a sampling distribution is normally shaped, we can apply the Empirical Rule and use Z-scores to determine probabilities. Let's look at some examples. Example A random sample of 100 students is taken from the population of all part-time...
A sample of 10 students record their scores on the final exam for their statistics class. The mean of the sample is 81 with sample standard deviation 7 points. Analysis of the 10 sample values indicated that the population is approximately normal. We wish to find the 95% confidence interval for the population mean test scores. What is the confidence level, c? Which of the following is correct? To find the confidence interval, a z-critical value should be used because...
Students in a statistics class are conducting a survey to estimate the mean pe numbers in the bones. number of units students at their college are enrolled in. The students collect a part es points random sample of 47 students. The mean of the sample is 12.3 units. The Part 2: 5 points sample has a standard deviation of 1.9 units. 20 points What is the 95% confidence interval for the average number of units that students in their college...
1. Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find the probability of obtaining a reading less than 1.089°C. P(Z<1.089)=P(Z<1.089)= (Round answer to four decimal places.) 2. Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is...
We can now use the Standard Normal Distribution Table to find the probability P(-0.25 sz s 1). 0.05 0.06 0.07 0.08 0.09 -0.2 0.4013 0.3974 0.3936 0.3897 0.3859 0.00 0.01 0.02 0.03 0.04 Using these 1.0 0.8413 0.8438 0.8461 0.8485 0.8531 The table entry for z = -0.25 is 0.00 and the table entry for z = 1 is values to calculate the probability gives the following result. PC-0.25 sz s 1) P(Z < 1) - P(Z 5 -0.25) 10....