It was recently reported that the heights of adults in Wyoming are normally distributed with a mean of 69 inches and a standard deviation of 1.5 inches. Round your answers to 4 decimal places. What is the probability that a randomly selected adult in Wyoming is less than 68.2 inches? Choose the correct probability notation. O P(x < 68.2) O P(ē < 68.2) What is the probability that a random sample of 13 adults from Wyoming have a mean height that is less than 68.2 inches? Choose the correct probability notation. O P(ã < 68.2) O P(x < 68.2)

Cody took his first physics exam and scored an 80. The population mean for this exam is 70, and the standard deviation is 5. What is the probability of selecting a person with a score greater than Cody’s?

Just do subpart "b"

The NJ Department of Health has reported the average life span of NJ residents is 81 years. If the standard deviation is σ = 4 years, what is the probability of a person living to at least 92 years of age?

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 $…

* Your answer is incorrect. Service time for a customer coming through a checkout counter in a retail store is a random variable with the mean of 3.5 minutes and standard deviation of 3.5 minutes. Suppose that the distribution of service time is fairly close to a normal distribution. Suppose there are two counters in a store, n₁ = 2 customers in the first line and n₂ = 13 customers in the second line. Find the probability that the difference between the mean service time for the shorter line X₁ and the mean service time for the longer one X₂ is more than 0.1 minutes. Assume that the service times for each customer can be regarded as independent random variables. Round your answer to two decimal places (e.g. 98.76). P = i 98.80

This is a 3 part question please help!