Express the confidence interval (58.2%,66%)(58.2%,66%) in the form of ˆp± E. _____% ± _______ %

Solve for the mean and standard deviation of the following Binomial distributions. a. n = 20 and p = 0.70 b. n = 70 and p = 0.35 c. n = 100 and p = 0.50

A researcher recorded the number of shots made, X, out of 10 free throws by a sample of 1000 randomly selected college students. The table below is a probability distribution table representing the data collected. Calculate the mean and the standard deviation of the probability distribution using a TI-83, TI-83 Plus, or TI-84 graphing calculator. x P(x) 0 0.01 1 0.01 2 0.02 3 0.04 4 0.03 5 0.05 6 0.21 7 0.29 8 0.21 9 0.08 10 0.05

A recent survey of 1040 U.S. adults selected at random showed that 634 consider the occupation of firefighter to have very great prestige. Estimate the probability (to the nearest hundredth) that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.Step 1Recall that the probability of an event based on relative frequency uses the formulaprobability of event = relative frequency = fn,where f is the frequency of the event occurrence in a sample of n observations. A total of 1040 people were surveyed and 634 considered the occupation of firefighter to have very great prestige. Therefore, the sample size is n =  . The event of interest is that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige. Therefore, the frequency f is equal to the number of adults who  the occupation of firefighter has very great prestige, so f =  (0.6096 answer is incorrect.) .

1:37 PM Wed Apr 21 * 72% 88 Q Stats Exam •.. Untitled Notebook (1) Stats Exam 37. The manager of Green Valley medical center claims that the median number of patients seen by doctors who work at the center is 85. To test this claim, 20 days are randomly selected and the number of patients seen is recorded as shown. At a=0.1, test the claim. (Use sign test) 86 82 90 88 85 85 94 93 69 61 89 90 91 94 78 85 71 77 69 77

Assuming a population is normally distributed, calculate the probability that a random member of the population falls between 3 standard deviations below the mean and 1 standard deviation below the mean. In other words in mathematical notation: P(-3 < z < -1) Enter your answer in decimal form. For example, 75% should be entered as 0.75.

The mean percent of childhood asthma prevalence in 43 cities is 2.13%. A random sample of 32 of these cities is selected. What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.5%? Interpret this probability. Assume that o = 1.34%. The probability is. (Round to four decimal places as needed.)