Find the standard deviation of random variable x

rd deviation of 3.2 hours. A sample of 35 high school seniors is selected. Find the following values, if possible. Please round your answers to four decimal places. If you should not find a probability, please write NA for your answer. What is the probability that the sample mean studying time is between 4 and 5 hours? What is the probability that the sample mean studying time is less than 6 hours? What is the probability that the sample mean studying time is more than 8.5 hours?

xt 4 tests are performed, find P(x=0), P(x=1), P(x=2),P(x=3) and P(x=4). SHOW ALL WORK TO GET CREDIT) 4) Use the confidence interval (99%) to solve the following question. A company is trying to establish the number of average amount of funds that are owed by its employees. They collect 1,000 accounts and found the sample average owed is $250 with a standard deviation of 10. Calculate the confidence interval (MUST SHOW ALL WORK TO GET CREDIT) 5) Use the t distribution formula and table to solve the following question. A random sample of 91 with a sample average of 90 and a standard deviation of 4.2 hours, calculate the confidence interval at 98% (MUST SHOW ALL WORK TO GET CREDIT) 6) A poll of 3,000 adults out of 5,500 was collected to found that they did not get a master’s degree. Calculate the confidence interval at 95%. (MUST SHOW ALL WORK TO GET CREDIT)

ion for the data. Thanks

illion dollars. If the mean revenue was 50 million dollars and the data has a standard deviation of 15 million, find the percentage. Assume that the distribution is normal. Round your answer to the nearest hundredth.

ard deviation of 18 million. Find the percentage of companies with revenue greater than 95 million dollars. Assume that the distribution is normal. Round your answer to the nearest hundredth.

n of μ= 160.4 and a standard deviation of σ = 9.0760. If all possible samples of size 13 are drawn from this population, find the percentage of them that would have means between 154 centimeters and 156 centimeters?

n of μ= 160.4 and a standard deviation of σ=9.0760. If all possible samples of size 13 are drawn from this population, find the percentage of them that would have means between 154 centimeters and 156 centimeters?

e customer is 4 per year with the standard deviation 2. Find the probability that the total number of yearly claims in the service department does not exceed 50000.

on with mean 81 and standard deviation 22. I have been asked to find the probability that X (bar) is greater than 86 and to find the 85th percentile of this data set

certain casino, the average amount lost by a random gambler is $20 per night, with a standard deviation of $135. If we randomly select 60 gamblers, find the probability that as a group they win more than they lose that night.dard deviation of $135. If we randomly select 60 gamblers, find the probability that as a group they win more than they lose that night.

deviation of $135. If we randomly select 60 gamblers, find the probability that as a group they win more than they lose that night.

consumes each year is 196 with a standard deviation of 22 pounds (Source: American Dietetic Association). If a sample of 50 individuals is randomly selected, find the probability that the mean of the sample will be less than 200 pounds.

ulation standard deviation is 4. Conduct the following test of hypothesis using the 0.05 significance level. H0: μ = 43 H1: μ ≠ 43 find the p value (round z value to 2 decimals and final answer 4 decimal place)

de the printout below. Use these values as estimates of the mean and standard deviation found in the population of all low prices. Suppose that the low prices were normally distributed (regardless of what your data may indicate). Find the proportion of all low prices that would be between $20 and $50 in the population. I want you to show your work. To receive full credit, you should include pictures of the normal curve (labeled with both x and z-values) with the pertinent probabilities shaded in the picture

f dieting: 2, 3, 1, 6, 2, 4 Assuming that these weights constitute an entire population, find the standard deviation of the population. Round your answer to two decimal places.

weighing between 140 lb and 191 lb. The new population of pilots has normally distributed weights with a mean of 145 lb and a standard deviation of 29.9 lb. a. If a pilot is randomly​ selected, find the probability that his weight is between 140 lb and 191 lb.

with a mean of 25 gm and a standard deviation of 5 gm. (a) If the machine is used 500 times, approximately how many times will it be expected to dispense 30 gm or more of chilli sauce? (b) How can you decrease this number to half? Give a numerical answer. 2. StarTech manufactures re sensors. They use a protective screen for their sensors to protect it from dust. The sensor becomes useless if the thickness of the screen exceeds 0.5 mm. They outsource the production of the screen to a di erent company that claims to manufacture screens with a mean thickness of 0.3 mm and a standard deviation of 0.1 mm. (a) If 10000 screens are manufactured how many will be discarded because they are too thick? (b) If screens less than 0.2 mm are too thin to be used, what is the probability that screens manufactured by the above company will be discarded because they are too thick or too thin? Show the result on a graph. 3. The amount of time that Sam spends playing the guitar is normally distributed with a mean of 15 hours and a standard deviation of 3 hours. (a) Find the probability that he spends between 15 and 18 hours playing the guitar during a given week. (b) What is the probability that he spends less than 3 hours playing the guitar during a given week? 4. Soon after he took oce in 1963, President Johnson was approved by 160 out of a sample of 200 Americans. With growing disillusionment over his Vietnam policy, by 1968 he was approved by only 70 out of a sample of 200 Americans. (a) What is the 90% con dence interval for the percentage of all Americans who approved of Johnson in 1963? In 1968? (b) What is the 90% con dence interval for the change?