The duration of a lunch break for one employee at a manufacturing company follows a normal distribution with mean µ minutes and standard deviation 5 minutes. The probability that a lunch break lasts for more than 52 minutes is 0.25. Find the value of µ. A. 55.37 B. 46.25 C. 57.75 D. 48.63
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Question 14
The duration of a lunch break for one employee at a manufacturing company follows a
distribution
lunch break lasts for more than 52 minutes is 0.25. Find the value of µ.
A. 55.37
B. 46.25
C. 57.75
D. 48.63
Step by step
Solved in 2 steps with 1 images
- Today, the waves are crashing onto the beach every 4.5 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 4.5 seconds. Round to 4 decimal places where possible. a. The mean of this distribution is b. The standard deviation is c. The probability that wave will crash onto the beach exactly 1.7 seconds after the person arrives is P(x = 1.7) = d. The probability that the wave will crash onto the beach between 1.5 and 3.9 seconds after the person arrives is P(1.5 1) = f. Find the maximum for the lower quartile. seconds.Answer number 2 &3Today, the waves are crashing onto the beach every 5.6 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 5.6 seconds. Round to 4 decimal places where possible. a. The mean of this distribution is b. The standard deviation is c. The probability that wave will crash onto the beach exactly 0.7 seconds after the person arrives is P(x = 0.7) = d. The probability that the wave will crash the beach between 1.6 and 5.1 seconds after the person arrives is P(1.6 3.72) = f. Suppose that the person has already been standing at the shoreline for 0.8 seconds without a wave crashing in. Find the probability that it will take between 1.4 and 3.3 seconds for the wave to crash onto the shoreline. g. 65% of the time a person will wait at least how long before the wave crashes in? seconds. h. Find the minimum for the upper quartile. seconds.
- Suppose that the walking step length of an adult males are normally distributed with a mean of 2.5ft and a standard deviation of 0.5ft. A sample of 50 man's step length is taken. Find the probability that the mean of the sample is taken less than 2.1ft. Round your answer to 4 decimal places if necessaryJ 1In a normal distribution, x = 5 and z = –1.4. This tells you that x = 5 is ____ standard deviations to the ____ (right orleft) of the mean.
- f the standard deviation of a normally distributed population is 22.0 and we take a sample of size 4, then the standard error of the mean isSuppose you take a random sample of n = 80 Douglas College students and ask each student their age. You find that the mean age of students in this sample is X = 22.0375 years and the sample standard deviation is s = 3.2014 years. Which of the following are sufficient conditions and are also true (to the best of our knowledge) for this study of Douglas College students’ ages? Select all that apply. A. X is normal variable B. n is greater or equal to 30 C. nq is greater or equal to 5 D. np is greater or equal to 5The mean amount of time it takes a kidney stone to pass is 14 days and the standard deviation is 6 days. Suppose that one individual is randomly chosen. Let X = time to pass the kidney stone. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X-N 14 ☑. 6 0° & b. Find the probability that a randomly selected person with a kidney stone will take longer than 10 days to pass it. c. Find the minimum number for the upper quarter of the time to pass a kidney stone. days. Submit Question
- Suppos are interested in studying a population to estimate its mean. The population is normal and has a standard deviation of G=23. We have taken a random sample of size 11=95 from the population. This is Sample 1 in the table below. (In the table, Sample 1 is written "S1", Sample 2 is written "S2", etc.) As shown in the table, the sample mean of Sample 1 is = 142.0. Also shown are the lower and upper limits of the 80% confidence interval for the population mean using this sample, as well as the lower and upper limits of the 95% confidence interval. Suppose that the true mean of the population is = 140, which is shown on the displays for the confidence intervals. Press the "Generate Samples" button to simulate taking 19 more random samples of size 11 = 95 from this same population. (The 80% and 95% confidence intervals for all of the samples are shown in the table and graphed.) Then complete parts (a) through (c) below the table. 80% 80% 95% 95% lower upper lower upper limit limit…A normal distribution has a mean of µ = 50 with σ = 30. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 50 and X = 80? Select one: a. 0.6826 b. 0.3413 c. 0.8413 d. 0.15873
