The output of a process is normally distributed with mean 19 and standard deviation 0.25. The...

The output of a process is normally distributed with mean 19 and standard deviation 0.25. The process specifications are at 20±1. What proportion of the process output is within the specifications? If the process target can be adjusted relatively easily, what would you recommend to production to reduce the proportion of nonconforming output?

Expert Solution

orchestra answered 2 years ago

Suppose that ? is normally distributed with mean 100 and standard deviation 19. A. What is...

Suppose that ? is normally distributed with mean 100 and standard deviation 19. A. What is the probability that ? is greater than 137.43? Probability = B. What value of ? does only the top 16% exceed?

a. A population is normally distributed with a mean of 16.4 and a standard deviation of...

a. A population is normally distributed with a mean of 16.4 and a standard deviation of 1.4. A sample of size 36 is taken from the population. What is the the standard deviation of the sampling distribution? Round to the nearest thousandth. b. A population is normally distributed with a mean of 15.7 and a standard deviation of 1.4. A sample of size 24 is taken from the population. What is the the standard deviation of the sampling distribution? Round...

If X is normally distributed with a mean of 30 and a standard deviation of 2,...

If X is normally distributed with a mean of 30 and a standard deviation of 2, find P(30 ≤ X ≤ 33.4). a) 0.455 b) 0.855 c) 0.955 d) 0.555 e) 0.755 f) None of the above

X is normally distributed with a mean of -9.74 and a standard deviation of 0.27. a....

X is normally distributed with a mean of -9.74 and a standard deviation of 0.27. a. Find the value of x that solves P(X ≤ x)= 0.25 b. Find the value of x that solves P(X > x)= 0.20 c. Find the value of x that solves P(x <X ≤ -9.70) = 0.20 d. Find the value of x that solves P(-9.74 < X ≤x) = 0.35

Suppose that X is normally distributed with a mean of 50 and a standard deviation of...

Suppose that X is normally distributed with a mean of 50 and a standard deviation of 12. What is P(X ≥ 74.96) a) 0.481 b) 0.019 c) 0.981 d) 0.024 e) 0.020 f) None of the above

Assume that X is normally distributed with a mean of 5 and a standard deviation of...

Assume that X is normally distributed with a mean of 5 and a standard deviation of 4. Determine the value for x P(-x < X-5 < x) = 0.99 Can someone please explain in detail how to solve it with the help of the Cumulative Normal Distribution Table? Thanks!

1. A population is normally distributed with a mean of 18 and a standard deviation of...

1. A population is normally distributed with a mean of 18 and a standard deviation of 2.5. What is the probability of randomly selecting one item from the population having: a) A value greater than 18. b) A value between 14 and 21.

Assume that a population is normally distributed with a mean of 100 and a standard deviation...

Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15. Would it be unusual for the mean of a sample of 3 to be 115 or more? Why or why not?

scores on the test are normally distributed with a mean of 76.4 and a standard deviation...

scores on the test are normally distributed with a mean of 76.4 and a standard deviation of 8.9. round answer nearest whole number. A. top 90% to 100% of scores b. scores below the top 90 % and above top 70% c. scores equal or below 70%and above the bottom 30% d. scores equal or below 30% snd above bottom 10%

IQ is normally distributed with a mean of 100 and a standard deviation of 15. a)...

IQ is normally distributed with a mean of 100 and a standard deviation of 15. a) Suppose one individual is randomly chosen. Find the probability that this person has an IQ greater than 95. Write your answer in percent form. Round to the nearest tenth of a percent. P (IQ greater than 95) = % b) Suppose one individual is randomly chosen. Find the probability that this person has an IQ less than 125. Write your answer in percent form....

Question