The probability distribution of the random variable X represents the number of hits a baseball player obtained in a game for the 2012 baseball season. 2 3 4 lo P(x) 0.1666 0.3292 0.2779 0.1492 0.0366 | 0.0405 The probability distribution was used along with statistical software to simulate 25 repetitions of the experiment (25 games). The number of hits was recorded. Approximate the mean and standard deviation of the random variable X based on the simulation. The simulation was repeated by performing 50 repetitions of the experiment. Approximate the mean and standard deviation of the random variable. Compare your results to the theoretical mean and standard deviation. What propert is being illustrated? E Click the icon to view the data tables. Compute the theoretical mean of the random variable X for the given probability distribution. Hx =O hits (Round to three decimal places as needed.)

The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a population mean of 60,000 miles and a population standard deviation of 4,000 miles. Suppose we select a sample of 90 tires and use a simulator to determine the tread life. What is the likelihood of finding that the sample mean is between 59,050 and 60,950? 0.4878 0.9756 0.9878 0.5122

Women's shoe sizes are (approximately) normally distributed, with an average size of 8.076 and a standard deviation of 1.468. You work at Vogue Magazine, and would like to interview some women about how shoes fit. You take a random sample of 10 women. What is the probability that the 10 women you sampled had an average shoe size that is less than 8?

In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 53.3 inches, and standard deviation of 2.2 inches.What is the probability that the height of a randomly chosen child is between 51.3 and 57.1 inches? Do not round until you get your your final answer, and then round to 3 decimal places.Answer=___________ (Round your answer to 3 decimal places.)

The median annual cost of auto insurance is $ 939 (According to DACO). Suppose the standard deviation is o = $ 241. What is the probability that in a simple random sample of auto insurance policies the sample mean differs more than $ 70.42 from the population mean if the sample size is 45? You must calculate the probability that the difference between the average cost in the sample and the average annual cost do not differ by $ 70.42. Find the probability that P (\ µ - š | 0) = P (-70.42 H-iS 70.42) Select one: a. The probability is 99% that the auto policy will not differ by $ 70.42 from the average annual cost. b. The probability is 50% that the auto policy will not differ by $ 70.42 from the average annual cost. c. The probability is 80% that the auto policy will not differ by $ 70.42 from the average annual cost. d. The probability is 70% that the auto policy will not differ by $ 70.42 from the average annual cost. e. The probability is 95% that the auto policy will not differ by $…

A binomial experiment is given. Decide whether you can use the normal distribution to approximate the binomial distribution. If you can, find the mean and standard deviation. If you cannot, explain why.  A survey of adults found that 44% have used a multivitamin in the past 12 months. You randomly select 50 adults and ask them if they have used a multivitamin in the past 12 months.