The 80th percentile is the value of X such that,
P(X < P80 ) = 0.80
For standard normal distribution,
P(Z < 0.8414) = 0.80
Hence for standard normal distribution, 0.8416 is the 80th percentile.
We know that
Question 9 (1 point) Suppose that replacement times for washing machine parts are normally distributed with...
Solve the problem. Suppose that replacement times for washing machines are normally distributed with a mean of 9.3 years and a standard deviation of 1.1 years. Find the probability that 70 randomly selected washing machines will have a mean replacement time less than 9.1 years. Write your answer as a decimal rounded to 4 places.
Suppose that replacement times for washing machines are normally distributed with a mean of 9.3 years and a standard deviation of 1.1 years. Find the replacement time that separates the top 3% from the bottom 97% . Round your answer to 3 decimal places.
Suppose that replacement times for washing machines are normally distributed with a mean of 11 years and a standard deviation of 19 years Find the replacement time that separates the top 18 from the botom 82% Round to the neareste Click to view.nage the love of the O A 12 years OB 127 years O C. 113 years OD 9 years
D Question 24 7 pts Suppose that replacement times for timing belts in cars are normally distributed with a mean of 9 years and a standard deviation of 3 years. Find the replacement time that separates the top 12% from the bottom 88%. (2 decimal places for the final answer) 7 pts
15) Assume that z scores are normally distributed with a mean of 0 and a standard deviation 15) of 1. If P(z> c) 0.109, find c. olve the problem. 16) 16) Scores on an English test are normally distributed with a mean of 37.4 and a standard deviation of 7.9. Find the score that separates the top 59% from the bottom 41% 17) Suppose that replacement times for washing machines are normally distributed with a 17) mean of 10.9 years...
Replacement times for televisions are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years. Find the probability that a randomly selected television will need a replacement time less than 6 years.
Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 10.6 years and a standard deviation of 1.2 years. If the company wants to provide a warranty so that only 3.8% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty? warranty = years Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores...
Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 10.6 years and a standard deviation of 1.2 years. If the company wants to provide a warranty so that only 1.3% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty? warranty = years Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores...
It is well documented that a typical washing machine can last anywhere between 5 to 20 years. Let the life of a washing machine be represented by a lognormal variable, Y = eX where X is normally distributed. In addition, let the mean and standard deviation of the life of a washing machine be 11 and half years and 2 years, respectively. [You may find it useful to reference the z table.] a. Compute the mean and the standard deviation...
It is well documented that a typical washing machine can last anywhere between 5 to 20 years. Let the life of a washing machine be represented by a lognormal variable, Y = eX where X is normally distributed. In addition, let the mean and standard deviation of the life of a washing machine be 9 and half years and 7 years, respectively. [You may find it useful to reference the z table.] a. Compute the mean and the standard deviation...